#### Graphing Absolute Value Functions

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###### Exercises
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Exercises 1 The parent function of all absolute value functions is: f(x)=∣x∣, and the general form of any absolute value function is: g(x)=a∣x−h∣+k. The added variables (a, h, and k) describe the placement of any and all transformations that can be made to the parent function. The coefficient a represents any shrinks, stretches, and reflections. The h represents any horizontal translations. And, the k represents any vertical translations. In the given equation, f(x)=-3∣x−1∣−4, the point (1,-4) is equal to (h,k) from the general form. This tells us how far the graph was shifted to the right and down. Because of the V-shaped nature of an absolute value function, this point is the center of the V which is called the vertex.
Exercises 2 A vertical stretch or shrink of a parent function can be written in the general form: y=a⋅f(x). In the case of an absolute value function, where f(x)=∣x∣, this is still the case but we have y=a∣x∣. So when looking at the general form for all absolute value functions, f(x)=a∣x−h∣+k, we still only need to be concerned about a when wondering if our function was shrunk or stretched. As with all vertical stretches and shrinks, when a>1 it's a stretch because you are creating larger outputs and when 0<a<1 it's a shrink because you are creating smaller outputs. Note, this is only the case because we are looking at vertical transformations.
Exercises 3 There are several types of transformations that can be done to absolute value functions. Let's look at the basics. We can start with the absolute value parent function, f(x)=∣x∣. Then we have:Reflection: g(x)=-∣x∣ Horizontal translation: g(x)=∣x−h∣ Vertical translation: g(x)=∣x∣+k Vertical stretch or shrink: g(x)=a∣x∣ Horizontal stretch or shrink:g(x)=∣ax∣
Exercises 4 To determine the graph of which function has the same y-intercept as the graph of f(x)=∣x−2∣+5, we will draw the three graphs, starting with f. To do so, we will make a table assigning some values to the x-variable and calculating their corresponding y-values.x∣x−2∣+5f(x)=∣x−2∣+5 -1∣-1−2∣+58 0∣0−2∣+57 1∣1−2∣+56 2∣2−2∣+55 3∣3−2∣+56 Now we will plot and connect the obtained points. Keep in mind that the graph of an absolute value function has a V shape.We see that the y-intercept of the graph is 7. We can graph g(x)=∣3x−2∣+5 and h(x)=3∣x−2∣+5 by following the same procedure.We see above that the graphs of g and f intercept the y-axis at the same point. Therefore, they have the same y-intercept.
Exercises 5 Let's graph d(x) first and then we can compare it to the graph of f(x).Graphing d(x) To graph the function, let's make a table of values first!x∣x∣−4Simplifyd(x) -5∣-5∣−45−41 -3∣-3∣−43−4-1 -1∣-1∣−41−4-3 0∣0∣−40−4-4 1∣1∣−41−4-3 3∣3∣−43−4-1 5∣5∣−45−41 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of d(x). Notice that d(x) is a transformation of f(x) and the graph of f(x)=∣x∣ is V-shaped. Thus, d(x) will also be a V-shaped graph.The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​ To find the range of an absolute value function, we need to think about where the vertex of the function is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The minimum of the given function is -4 and then it will continue increasing indefinitely. Range:  -4≤y<∞​Comparing the Functions To compare our graph to the graph f(x)=∣x∣, let's draw them on one coordinate plane.As we can see, the graph of d(x) is a vertical translation 4 units down of the graph f(x).
Exercises 6 Let's graph r(x) first and then we can compare it to the graph of f(x).Graphing r(x) To graph the function, let's make a table of values first!x∣x∣+5Simplifyr(x) -5∣-5∣+55+510 -3∣-3∣+53+58 -1∣-1∣+51+56 0∣0∣+50+55 1∣1∣+51+56 3∣3∣+53+58 5∣5∣+55+510 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of r(x). Notice that r(x) is a transformation of f(x) and the graph of f(x)=∣x∣ is V-shaped. Thus, r(x) will also be a V-shaped graph.The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​ To find the range of an absolute value function, we need to think about where the vertex of the function is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The minimum of the given function is 5 and then it will continue increasing indefinitely. Range: 5≤y<∞​Comparing the Functions To compare our graph to the graph f(x)=∣x∣, let's draw them on one coordinate plane.As we can see, the graph of r(x) is a vertical translation 5 units up of the graph f(x).
Exercises 7 Let's graph m(x) first and then we can compare it to the graph of f(x).Graphing m(x) To graph the function, let's make a table of values first!x∣x+1∣Simplifym(x) -7∣-7+1∣∣-6∣6 -5∣-5+1∣∣-4∣4 -3∣-3+1∣∣-2∣2 -1∣-1+1∣∣0∣0 1∣1+1∣∣2∣2 3∣3+1∣∣4∣4 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of m(x). Notice that m(x) is a transformation of f(x) and the graph of f(x)=∣x∣ is V-shaped. Thus, m(x) will also be a V-shaped graph.The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​ To find the range of an absolute value function, we need to think about where the vertex of the function is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The minimum of the given function is 0 and then it will continue increasing indefinitely. Range: 0≤y<∞​Comparing the Functions To compare our graph to the graph f(x)=∣x∣, let's draw them on one coordinate plane.As we can see, the graph of m(x) is a horizontal translation 1 unit left of the graph f(x).
Exercises 8 Let's graph v(x) first and then we can compare it to the graph of f(x).Graphing v(x) To graph the function, let's make a table of values first!x∣x−3∣Simplifyv(x) -1∣-1−3∣∣-4∣4 1∣1−3∣∣-2∣2 3∣3−3∣∣0∣0 5∣5−3∣∣2∣2 7∣7−3∣∣4∣4 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of v(x). Notice that v(x) is a transformation of f(x) and the graph of f(x)=∣x∣ is V-shaped. Thus, v(x) will also be a V-shaped graph.The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​ To find the range of an absolute value function, we need to think about where the vertex of the function is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The minimum of the given function is 0 and then it will continue increasing indefinitely. Range: 0≤y<∞​Comparing the Functions To compare our graph to the graph f(x)=∣x∣, let's draw them on one coordinate plane.As we can see, the graph of v(x) is a horizontal translation 3 units right of the graph f(x).
Exercises 9 Let's graph p(x) first and then we can compare it to the graph of f(x).Graphing p(x) To graph the function, let's make a table of values first!x31​∣x∣Simplifyp(x) -931​∣-9∣31​(9)3 -631​∣-6∣31​(6)2 -331​∣-3∣31​(3)1 031​∣0∣31​(0)0 331​∣3∣31​(3)1 631​∣6∣31​(6)2 931​∣9∣31​(9)3 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of p(x). Notice that p(x) is a transformation of f(x) and the graph of f(x)=∣x∣ is V-shaped. Thus, p(x) will also be a V-shaped graph.The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​ To find the range of an absolute value function, we need to think about where the vertex of the function is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The minimum of the given function is 0 and then it will continue increasing indefinitely. Range: 0≤y<∞​Comparing the Functions To compare our graph to the graph f(x)=∣x∣, let's draw them on one coordinate plane.As we can see, the graph of p(x) is a vertical shrink of the graph f(x) by a factor of 31​.
Exercises 10 Let's graph j(x) first and then we can compare it to the graph of f(x).Graphing j(x) To graph the function, let's make a table of values first!x3∣x∣Simplifyj(x) -33∣-3∣3(3)9 -23∣-2∣3(2)6 -13∣-1∣3(1)3 03∣0∣3(0)0 13∣1∣3(1)3 23∣2∣3(2)6 33∣3∣3(3)9 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of j(x). Notice that j(x) is a transformation of f(x) and the graph of f(x)=∣x∣ is V-shaped. Thus, j(x) will also be a V-shaped graph.The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​ To find the range of an absolute value function, we need to think about where the vertex of the function is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The minimum of the given function is 0 and then it will continue increasing indefinitely. Range: 0≤y<∞​Comparing the Functions To compare our graph to the graph f(x)=∣x∣, let's draw them on one coordinate plane.As we can see, the graph of j(x) is a vertical stretch of the graph f(x) by a factor of 3.
Exercises 11 Let's graph a(x) first and then we can compare it to the graph of f(x).Graphing a(x) To graph the function, let's make a table of values first!x-5∣x∣Simplifya(x) -2-5∣-2∣-5(2)-10 -1-5∣-1∣-5(1)-5 0-5∣0∣-5(0)0 1-5∣1∣-5(1)-5 2-5∣2∣-5(2)-10 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of a(x). Notice that a(x) is a transformation of f(x) and the graph of f(x)=∣x∣ is V-shaped. Thus, a(x) will also be a V-shaped graph.The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​ To find the range of an absolute value function, we need to think about where the vertex of the function is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The maximum of the given function is 0 and then it will continue decreasing indefinitely. Range: -∞<y≤0​Comparing the Functions To compare our graph to the graph f(x)=∣x∣, let's draw them on one coordinate plane.As we can see, the graph of a(x) is a vertical stretch of the graph f(x) by a factor of 5 and a reflection in the x-axis.
Exercises 12 Let's graph q(x) first and then we can compare it to the graph of f(x).Graphing q(x) To graph the function, let's make a table of values first!x-23​∣x∣Simplifyq(x) -4-23​∣-4∣-23​(4)-6 -2-23​∣-2∣-23​(2)-3 0-23​∣0∣-23​(0)0 2-23​∣-2∣-23​(2)-3 4-23​∣4∣-23​(4)-6 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of q(x). Notice that q(x) is a transformation of f(x) and the graph of f(x)=∣x∣ is V-shaped. Thus, q(x) will also be a V-shaped graph.The domain of an absolute value function will usually be all real numbers, unless specific restrictions have been imposed upon the function. Domain: -∞<x<∞​ To find the range of an absolute value function, we need to think about where the vertex of the function is located. Because this type of function will always have the same basic V-shape, the y-value of the vertex is the minimum or maximum of the range. The maximum of the given function is 0 and then it will continue decreasing indefinitely. Range: -∞<y≤0​Comparing the Functions To compare our graph to the graph f(x)=∣x∣, let's draw them on one coordinate plane.As we can see, the graph of q(x) is a vertical stretch of the graph f(x) by a factor of 23​ and a reflection in the x-axis.
Exercises 13 Let's graph h(x) first and then we can compare it to the graph of f(x).Graphing h(x) To graph the function, let's make a table of values first!x∣x−6∣+2Simplifyh(x) 0∣0−6∣+2∣-6∣+28 2∣2−6∣+2∣-4∣+26 4∣4−6∣+2∣-2∣+24 6∣6−6∣+2∣0∣+22 8∣8−6∣+2∣2∣+24 10∣10−6∣+2∣4∣+26 12∣12−6∣+2∣6∣+28 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of h(x). Notice that h(x) is a transformation of the graph of the parent function y=∣x∣, which is V-shaped. Thus, h(x) will also be a V-shaped graph.Now that we have graphed our function, we can move on to comparing it to f(x).Comparing the Functions To compare the graph of h(x) to the graph of f(x)=∣x−6∣, let's first draw this function. Notice that f(x) is a horizontal translation 6 units to the right of the graph of the parent function y=∣x∣.Now that we have already drawn f(x), we can compare h(x) with it.As we can see, the graph of h(x) is a vertical translation 2 units up from the graph of f(x).
Exercises 14 Let's graph n(x) first and then we can compare it to the graph of f(x).Graphing h(x) To graph the function, let's make a table of values first!x21​∣x−6∣Simplifyn(x) 021​∣0−6∣21​∣-6∣3 221​∣2−6∣21​∣-4∣2 421​∣4−6∣21​∣-2∣1 621​∣6−6∣21​∣0∣0 821​∣8−6∣21​∣2∣1 1021​∣10−6∣21​∣4∣2 1221​∣12−6∣21​∣6∣3 We plot these ordered pairs on a coordinate plane and connect them to get the graph of n(x). Notice that n(x) is a transformation of the parent function y=∣x∣, which is V-shaped. Thus, n(x) will also be a V-shaped graph.As we graphed our function we can move on to comparing it to f(x).Comparing the Functions To compare the graph of n(x) to the graph of f(x)=∣x−6∣, let's first draw this function. Notice that f(x) is a horizontal translation 6 units right of the graph of y=∣x∣.Now that we have already drawn f(x), we can compare it to n(x).As we can see, the graph of n(x) is a vertical shrink of the graph f(x) by a factor of 21​.
Exercises 15 Let's graph k(x) first and then we can compare it to the graph of f(x).Graphing k(x) To graph the function, let's make a table of values first!x-3∣x−6∣Simplifyk(x) 3-3∣3−6∣-3∣-3∣-9 4-3∣4−6∣-3∣-2∣-6 5-3∣5−6∣-3∣-1∣-3 6-3∣6−6∣-3∣0∣0 7-3∣7−6∣-3∣1∣-3 8-3∣8−6∣-3∣2∣-6 9-3∣9−6∣-3∣3∣-9 We plot these ordered pairs on a coordinate plane and connect them to get the graph of k(x). Notice that k(x) is a transformation of the graph of the parent function y=∣x∣, which is V-shaped. Thus, k(x) will also be a V-shaped graph.Now we can move on to comparing this function with f(x).Comparing the Functions To compare the graph of k(x) to the graph f(x)=∣x−6∣, let's first draw this function. Notice that f(x) is a horizontal translation 6 units right of the graph of y=∣x∣.Now that we have drawn f(x), we can compare it to k(x).As we can see, the graph of k(x) is a vertical stretch of the graph f(x) by a factor of 3 and a reflection in the x-axis.
Exercises 16 Let's graph g(x) first and then we can compare it to the graph of f(x).Graphing g(x) To graph the function, let's make a table of values first!x∣x−1∣Simplifyg(x) -3∣-3−1∣∣-4∣4 -1∣-1−1∣∣-2∣2 1∣1−1∣∣0∣0 3∣3−1∣∣2∣2 5∣5−1∣∣4∣4 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of g(x). Notice that g(x) is a transformation of the graph of the parent function y=∣x∣, which is V-shaped. Thus, g(x) will also be a V-shaped graph.As we graphed our function we can move on to comparing it with function f.Comparing the Functions To compare our graph to the graph f(x)=∣x−6∣, let's first draw this function. Notice that this function is a horizontal translation 6 units right of the graph y=∣x∣.Now when we have already drawn function f we can compare g(x) with it.As we can see, the graph of g(x) is a horizontal translation 5 units left of the graph f(x).
Exercises 17 Let's graph y(x) first and then we can compare it to the graph of f(x).Graphing y(x) To graph the function, let's make a table of values first!x∣x+4∣−2Simplifyy(x) -8∣-8+4∣−2∣-4∣−22 -6∣-6+4∣−2∣-2∣−20 -4∣-4+4∣−2∣0∣−2-2 -2∣-2+4∣−2∣2∣−20 0∣0+4∣−2∣4∣−22 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of y(x). Notice that y(x) is a transformation of the graph of the parent function y=∣x∣, which is V-shaped. Thus, y(x) will also be a V-shaped graph.As we graphed our function we can move on to comparing it with function f.Comparing the Functions To compare our graph to the graph f(x)=∣x+3∣−2, let's first draw this function. We will start with making a table of values just like with function y.x∣x+3∣−2Simplifyf(x) -7∣-7+3∣−2∣-4∣−22 -5∣-5+3∣−2∣-2∣−20 -3∣-3+3∣−2∣0∣−2-2 -1∣-1+3∣−2∣2∣−20 1∣1+3∣−2∣4∣−22Now when we have already drawn function f we can compare g(x) with it.As we can see, the graph of y(x) is a horizontal translation 1 unit left of the graph f(x).
Exercises 18 Let's graph b(x) first and then we can compare it to the graph of f(x).Graphing b(x) To graph the function, let's make a table of values first!x∣x+3∣+3Simplifyb(x) -7∣-7+3∣+3∣-4∣+37 -5∣-5+3∣+3∣-2∣+35 -3∣-3+3∣+3∣0∣+33 -1∣-1+3∣+3∣2∣+35 1∣1+3∣+3∣4∣+37 Now we can plot these ordered pairs on a coordinate plane and connect them to get the graph of b(x). Notice that b(x) is a transformation of the graph of the parent function y=∣x∣, which is V-shaped. Thus, b(x) will also be a V-shaped graph.As we graphed our function we can move on to comparing it with function f.Comparing the Functions To compare our graph to the graph f(x)=∣x+3∣−2, let's first draw this function. We will start with making a table of values just like with function y.x∣x+3∣−2Simplifyf(x) -7∣-7+3∣−2∣-4∣−22 -5∣-5+3∣−2∣-2∣−20 -3∣-3+3∣−2∣0∣−2-2 -1∣-1+3∣−2∣2∣−20 1∣1+3∣−2∣4∣−22Now when we have already drawn function f we can compare b(x) with it.As we can see, the graph of b(x) is a vertical translation 5 units up of the graph f(x).
Exercises 19 The graph of g(x) is 3 units lower than f(x), this is a vertical translation by 3 units down. We can see this by looking at the graphs.As a result, the value of k is -3. g(x)=∣x∣−3​
Exercises 20 When we look at the graphs, we can see that t(x) has the same shape as f(x), but it is shifted 1 unit to the right. This is a horizontal translation.As a result, the value of h is 1. t(x)=∣x−1∣​
Exercises 21 The graphs of f(x)=∣x∣ and p(x)=a∣x∣ both have vertices located at the origin, so we know that there haven't been any translations. However, p(x) is upside down and somewhat stretched. The fact that p(x) is upside down means that it has been reflected in the x-axis. This means a is a negative number. a<0​ Let's assume, temporarily, that a=-1. In this case, we would only change the sign of the graph's y-values, taking us from the graph of f(x)=∣x∣ to the graph of y=-∣x∣, shown below.To get from the graph of y=-∣x∣ to the graph of p(x), we need to stretch y=-∣x∣ away from the x-axis. Let's look at a point on each graph with the same x-values and compare their y-values.The graph of y=-∣x∣ passes through (1,-1) and the graph of p(x) passes through (1,-3). This means that the y-values in y=-∣x∣ have been stretched by a factor of 3. y=-∣x∣y-value​-1​​×∣factorstretch​3​​=p(x)=a∣x∣y-value​-3​​​ Therefore, to get both a reflection in the x-axis and a stretch of 3, we must have that a=-3.
Exercises 22 The graphs of f(x)=∣x∣ and w(x)=a∣x∣ both have vertices located at the origin, so we know that there haven't been any translations. However, w(x) is somewhat shrunk. This means a is less than 1. a<1​ To get from the graph of f(x) to the graph of w(x), we need to shrink f away from the y-axis. Let's look at a point on each graph with the same y-values and compare their x-values.As we can see from the graph, when the x-value of w(x) is two times the x-value of f(x), the y-value is the same for both functions. Therefore, to get w(x), f(x) must be vertically shrunk by a factor of a=21​.
Exercises 23 We are asked to translate the graph of g(x)=∣x∣ down 7 units. Recall that, when translating absolute value functions, vertical translations occur outside the absolute value symbol. Let's consider a general equation, where a is a real number. y=∣x∣+a​ The graph of the above equation is a vertical translation up a units of the graph of g(x)=∣x∣. In our case, since the translation is performed down, the value of a is -7. Let h(x) be the function whose graph represents the desired translation. h(x)=∣x∣+(-7)⇔h(x)=∣x∣−7​
Exercises 24 The graph of y=∣x−a∣ is the general form for a horizontal translation of y=∣x∣.If a is a positive number, then y=∣x−a∣ translates the graph of y=∣x∣ horizontally to the right by a units. If a is a negative number, it is a translation to the left. We can use this to create the equation for the translation of y=∣x∣ to the left by 10 units. y=∣x−(-10)∣⇔y=∣x+10∣​
Exercises 25 The graph of y=a∣x∣ is the general form for a vertical stretch or shrink of y=∣x∣ by the scale factor a.If a>1, then y=a∣x∣ moves the points of the graph of y=∣x∣ further away from the x-axis. We have a stretch. If 0<a<1, then y=a∣x∣ moves the points of the graph of y=∣x∣ closer to the x-axis. We have a shrink.We can use this to create the equation for shrinking y=∣x∣ by a scale factor 41​. y=41​∣x∣​
Exercises 26 The graph of y=a∣x∣ is the general form for a vertical stretch or shrink of y=∣x∣ by the scale factor a.If a>1, then y=a∣x∣ moves the points of the graph of y=∣x∣ further away from the x-axis. We have a stretch. If 0<a<1 then y=a∣x∣ moves the points of the graph of y=∣x∣ closer to the x-axis. We have a shrink.We can use this to write the equation for stretching y=∣x∣ by a factor of 3. y=3∣x∣​ Now, if we want to also reflect our function across the x-axis, we need each and every y-value to change signs.We can do this by multiplying the outputs (y-values) by -1. In general, this is written in the form -f(x). We can, however, write this to better fit the given exercise. y=-3∣x∣​
Exercises 27 To graph the functions without going through the entire process of transforming the parent function, we can make two tables of values. Then, we only need to plot and connect the obtained points.Graph Let's begin with creating a table of values for f.x∣x−4∣f(x)=∣x−4∣ -1∣-1−4∣5 1∣1−4∣3 3∣3−4∣1 5∣5−4∣1 Now it is time of doing the same with function g.x∣3x−4∣g(x)=∣3x−4∣ -1∣3(-1)−4∣7 1∣3(1)−4∣1 3∣3(3)−4∣5 5∣3(5)−4∣11 Finally, we will plot and connect these ordered pairs. Keep in mind that the graph of an absolute value function has a V shape!Comparison Note that we can rewrite g in terms of f. f(x)=∣x−4∣g(x)=∣3x−4∣​⇒g(x)=f(3x)​ The graph of g is a horizontal shrink of the graph of f by a factor of 31​. The y-intercept is the same for both graphs. When the input values of f are three times the input values of g, the output values of f and g are the same.
Exercises 28 To graph the functions without going through the entire process of transforming the parent function, we can make two tables of values. Then we only need to plot the ordered pairs that we find.Graph Let's begin with creating a table of values of function h.x∣x+5∣Simplifyh(x) -5∣-5+5∣∣0∣0 -3∣-3+5∣∣2∣2 -1∣-1+5∣∣4∣4 1∣1+5∣∣6∣6 Now it is time for doing the same with function t.x∣2x+5∣Simplifyt(x) -5∣2(-5)+5∣∣-5∣5 -3∣2(-3)+5∣∣-1∣1 -1∣2(-1)+5∣∣3∣3 1∣2(1)+5∣∣7∣7 Finally we can take these ordered pairs and plot them on one coordinate plane.Comparison We can rewrite t(x) as h(2x) which is a horizontal shrink by a factor of 21​. It is being shrunk closer the y-axis. Notice that when the x-values of f are two times the x-values of g the y-values stay the same.
Exercises 29 To graph the functions without going through the entire process of transforming the parent function, we can make two tables of values. Then we only need to plot the ordered pairs that we find.Graph Let's begin with creating a table of values of function p.x∣x+1∣−2Simplifyp(x) -5∣-5+1∣−2∣-4∣−22 -3∣-3+1∣−2∣-2∣−20 -1∣-1+1∣−2∣0∣−2-2 1∣1+1∣−2∣2∣−20 3∣3+1∣−2∣4∣−22 Now it is time for doing the same with function q.x∣∣∣∣​41​x+1∣∣∣∣​−2Simplifyq(x) -5∣∣∣∣​41​(-5)+1∣∣∣∣​−2∣∣∣∣​-41​∣∣∣∣​−2-143​ -3∣∣∣∣​41​(-3)+1∣∣∣∣​−2∣∣∣∣​41​∣∣∣∣​−2-143​ -1∣∣∣∣​41​(-1)+1∣∣∣∣​−2∣∣∣∣​43​∣∣∣∣​−2-141​ 1∣∣∣∣​41​(1)+1∣∣∣∣​−2∣∣​141​∣∣​−2-43​ 3∣∣∣∣​41​(3)+1∣∣∣∣​−2∣∣​143​∣∣​−2-41​ Finally we can take these ordered pairs and plot them on one coordinate plane.Comparison We can rewrite q(x) as p(41​x) which is a horizontal stretch by a factor of 4. It is being stretched farther away from the y-axis. Note, while this may look like a horizontal translation, it is not. The x-intercept changes not because the graph was shifted but because the input values are being moved four times as far away from the y-axis.
Exercises 30 To graph the functions without going through the entire process of transforming the parent function, we can make two tables of values. Then we only need to plot the ordered pairs that we find.Graph Let's begin with creating a table of values of function w.x∣x−3∣+4Simplifyw(x) -2∣-2−3∣+4∣-5∣+49 0∣0−3∣+4∣-3∣+47 2∣2−3∣+4∣-1∣+45 4∣4−3∣+4∣1∣+45 Now it is time for doing the same with function y.x∣5x−3∣+4Simplifyy(x) -1∣5(-1)−3∣+4∣-8∣+412 53​∣∣∣∣​5(53​)−3∣∣∣∣​+4∣0∣+44 1∣5(1)−3∣+4∣2∣+46 2∣5(2)−3∣+4∣7∣+411 Finally we can take these ordered pairs and plot them on one coordinate plane.Comparison We can rewrite y(x) as w(5x) which is a horizontal shrink by a factor of 51​. It is being shrunk closer the y-axis. Notice that when the x-values of w are five times the x-values of y the y-values stay the same.
Exercises 31 To graph the functions without going through the entire process of transforming the parent function, we can make two tables of values. Then we only need to plot and connect the obtained points.Graph Let's begin with creating a table of values for a.x∣x+2∣+3a(x) -3∣-3+2∣+34 -1∣-1+2∣+34 1∣1+2∣+36 3∣3+2∣+38 Now it is time for doing the same with function b.x∣-4x+2∣+3b(x) -3∣-4(-3)+2∣+317 -1∣-4(-1)+2∣+39 1∣-4(1)+2∣+35 3∣-4(3)+2∣+313 Finally, we will plot and connect these ordered pairs. Keep in mind that the graph of an absolute value function has a V shape!Comparison Note that we can rewrite b in terms of a. a(x)=∣x+2∣+3b(x)=∣-4x+2∣+3​⇒b(x)=a(-4x)​ The graph of b is a horizontal shrink of the graph of a by a factor of 41​, followed by a reflection in the y-axis. Notice that the y-intercept is the same for both graphs.
Exercises 32 To graph the functions without going through the entire process of transforming the parent function, we can make two tables of values. Then we only need to plot the ordered pairs that we find.Graph Let's begin with creating a table of values of function u.x∣x−1∣+2Simplifyu(x) -3∣-3−1∣+2∣-4∣+26 -1∣-1−1∣+2∣-2∣+24 1∣1−1∣+2∣0∣+22 3∣3−1∣+2∣2∣+24 Now it is time for doing the same with function v.x∣∣∣∣​-21​x−1∣∣∣∣​+2Simplifyv(x) -4∣∣∣∣​-21​(-4)−1∣∣∣∣​+2∣1∣+23 -2∣∣∣∣​-21​(-2)−1∣∣∣∣​+2∣0∣+22 2∣∣∣∣​-21​(2)−1∣∣∣∣​+2∣-2∣+24 4∣∣∣∣​-21​(4)−1∣∣∣∣​+2∣-3∣+25 Finally we can take these ordered pairs and plot them on one coordinate plane.Comparison We can rewrite v(x) as u(-21​x) which is a horizontal stretch by a factor of 2 and next the reflection in the y-axis. Notice that when the y-values stay the same.
Exercises 33 First we will go through all the transformations from the graph of f(x)=∣x∣ to the graph of given function. After that we will make a table of values to graph this function.Transformations We can look at each of the transformations individually and work our way from the "inside" to the "outside." r(x)=∣x+2∣−6​ The given equation is a transformation of the parent function f(x)=∣x∣. We can define each transformation from the inside to the outside. Start with the transformation closest to x and end with the farthest from x. First there is a horizontal translation 2 units to the left, ∣x∣⇒∣x+2∣. Remember, typically ∣x−h∣ is a horizontal translation h units to the right, while ∣x+h∣ is translated h units to the left.Next, we have a vertical translation 6 units down, ∣x+2∣⇒∣x+2∣−6. Remember, in most cases ∣x∣+k is a vertical translation k units up, while ∣x∣−k is translated k units down.Graphing r(x) To graph the function without going through the entire process of transforming the parent function, we can make a table of values. Then we only need to plot the ordered pairs that we find.x∣x+2∣−6Simplifyr(x) -6∣-6+2∣−6∣-4∣−6-2 -4∣-4+2∣−6∣-2∣−6-4 -2∣-2+2∣−6∣0∣−6-6 0∣0+2∣−6∣2∣−6-4 2∣2+2∣−6∣4∣−6-2 4∣4+2∣−6∣6∣−60 Now we can plot these points and connect them to create our graph of r(x).
Exercises 34 First we will go through all the transformations from the graph of f(x)=∣x∣ to the graph of given function. After that we will make a table of values to graph this function.Transformations We can look at each of the transformations individually and work our way from the "inside" to the "outside." c(x)=∣x+4∣+4​ The given equation is a transformation of the parent function f(x)=∣x∣. We can define each transformation from the inside to the outside. Start with the transformation closest to x and end with the farthest from x. First there is a horizontal translation 4 units to the left, ∣x∣⇒∣x+4∣. Remember, typically ∣x−h∣ is a horizontal translation h units to the right, while ∣x+h∣ is translated h units to the left.Next, we have a vertical translation 4 units up, ∣x+4∣⇒∣x+4∣+4. Remember, in most cases ∣x∣+k is a vertical translation k units up, while ∣x∣−k is translated k units down.Graphing c(x) To graph the function without going through the entire process of transforming the parent function, we can make a table of values. Then we only need to plot the ordered pairs that we find.x∣x+4∣+4Simplifyc(x) -8∣-8+4∣+4∣-4∣+48 -6∣-6+4∣+4∣-2∣+46 -4∣-4+4∣+4∣0∣+44 -2∣-2+4∣+4∣2∣+46 0∣0+4∣+4∣4∣+48 Now we can plot these points and connect them to create our graph of r(x).
Exercises 35 First we will go through all the transformations from the graph of f(x)=∣x∣ to the graph of given function. After that we will make a table of values to graph this function.Transformations We can look at each of the transformations individually and work our way from the "inside" to the "outside." d(x)=-∣x−3∣+5​ The given equation is a transformation of the parent function f(x)=∣x∣. We can define each transformation from the inside to the outside. Start with the transformation closest to x and end with the farthest from x. First there is a horizontal translation 3 units to the right, ∣x∣⇒∣x−3∣. Remember, typically ∣x−h∣ is a horizontal translation h units to the right, while ∣x+h∣ is translated h units to the left.Next we have a reflection in the x-axis, ∣x−3∣⇒-∣x−3∣.Finally, we have a vertical translation 5 units up, -∣x−3∣⇒-∣x−3∣+5.Graphing d(x) To graph the function without going through the entire process of transforming the parent function, we can make a table of values. Then we only need to plot the ordered pairs that we find.</listcircle>x-∣x−3∣+5Simplifyd(x) -1-∣-1−3∣+5-∣-4∣+51 1-∣1−3∣+5-∣-2∣+53 3-∣3−3∣+5-∣0∣+55 5-∣5−3∣+5-∣2∣+53 7-∣7−3∣+5-∣4∣+51 Now we can plot these points and connect them to create our graph of d(x).
Exercises 36 First we will go through all the transformations from the graph of f(x)=∣x∣ to the graph of given function. After that we will make a table of values to graph this function.Transformations We can look at each of the transformations individually and work our way from the "inside" to the "outside." v(x)=-3∣x+1∣+4​ The given equation is a transformation of the parent function f(x)=∣x∣. We can define each transformation from the inside to the outside. Start with the transformation closest to x and end with the farthest from x. First there is a horizontal translation 1 units to the left, ∣x∣⇒∣x+1∣. Remember, typically ∣x−h∣ is a horizontal translation h units to the right, while ∣x+h∣ is translated h units to the left.Then we have a vertical stretch by a factor of 3, ∣x+1∣⇒3∣x+1∣.Next we have a reflection in the x-axis, 3∣x+1∣⇒-3∣x+1∣.Finally, we have a vertical translation 4 units up, -3∣x+1∣⇒-3∣x+1∣+4.Graphing v(x) To graph the function without going through the entire process of transforming the parent function, we can make a table of values. Then we only need to plot the ordered pairs that we find.x-3∣x+1∣+4Simplifyv(x) -4-3∣-4+1∣+4-3∣-3∣+4-5 -3-3∣-3+1∣+4-3∣-2∣+4-2 -2-3∣-2+1∣+4-3∣-1∣+41 -1-3∣-1+1∣+4-3∣0∣+44 0-3∣0+1∣+4-3∣1∣+41 1-3∣1+1∣+4-3∣2∣+4-2 Now we can plot these points and connect them to create our graph of g(x).
Exercises 37 First we will go through all the transformations from the graph of f(x)=∣x∣ to the graph of given function. After that we will make a table of values to graph this function.Transformations We can look at each of the transformations individually and work our way from the "inside" to the "outside." m(x)=∣∣∣∣​21​x+4∣∣∣∣​−1​ The given equation is a transformation of the parent function f(x)=∣x∣. We can define each transformation from the inside to the outside. Start with the transformation closest to x and end with the farthest from x. First there is a horizontal translation 4 units to the left, ∣x∣⇒∣x+4∣. Remember, typically ∣x−h∣ is a horizontal translation h units to the right, while ∣x+h∣ is translated h units to the left.After that we have a vertical shrink by a factor of 21​, ∣x∣⇒21​∣x∣.Finally, we have a vertical translation 1 unit down, 21​∣x+4∣⇒21​∣x+4∣−1.Graphing m(x) To graph the function without going through the entire process of transforming the parent function, we can make a table of values. Then we only need to plot the ordered pairs that we find.x21​∣x+4∣−1Simplifym(x) -1221​∣-12+4∣−121​∣-8∣−13 -821​∣-8+4∣−121​∣-4∣−11 -421​∣-4+4∣−121​∣0∣−1-1 021​∣0+4∣−121​∣4∣−11 421​∣4+4∣−121​∣8∣−13 Now we can plot these points and connect them to create our graph of m(x).
Exercises 38 First we will go through all the transformations from the graph of f(x)=∣x∣ to the graph of given function. After that we will make a table of values to graph this function.Transformations We can look at each of the transformations individually and work our way from the "inside" to the "outside." s(x)=∣2x−2∣−3​ The given equation is a transformation of the parent function f(x)=∣x∣. We can define each transformation from the inside to the outside. Start with the transformation closest to x and end with the farthest from x. First we have a horizontal shrink by a factor of 21​, ∣x∣⇒∣2x∣.Then there is a horizontal translation 2 units to the right, ∣2x∣⇒∣2x−2∣. Remember, typically ∣x−h∣ is a horizontal translation h units to the right, while ∣x+h∣ is translated h units to the left.Finally, we have a vertical translation 3 units down, ∣2x−2∣⇒∣2x−2∣−3.Graphing s(x) To graph the function without going through the entire process of transforming the parent function, we can make a table of values. Then we only need to plot the ordered pairs that we find.x∣2x−2∣−3Simplifys(x) -2∣2(-2)−2∣−3∣-6∣−33 -1∣2(-1)−2∣−3∣-4∣−31 0∣2(0)−2∣−3∣-2∣−3-1 1∣2(1)−2∣−3∣0∣−3-3 2∣2(2)−2∣−3∣2∣−3-1 3∣2(3)−2∣−3∣4∣−31 Now we can plot these points and connect them to create our graph of s(x).
Exercises 39 First we will go through all the transformations from the graph of f(x)=∣x∣ to the graph of given function. After that we will make a table of values to graph this function.Transformations We can look at each of the transformations individually and work our way from the "inside" to the "outside." j(x)=∣-x+1∣−5​ The given equation is a transformation of the parent function f(x)=∣x∣. We can define each transformation from the inside to the outside. Start with the transformation closest to x and end with the farthest from x. First we have a reflection in the y-axis, ∣x∣⇒∣-x∣. Notice that, since the graph is symmetric this will not change how it looks.Then there is a horizontal translation 1 unit to the left, ∣-x∣⇒∣-x+1∣. However, since our function has been reflected with respect to the vertical axis, it will be translated to the right instead. Remember, typically ∣x−h∣ is a horizontal translation h units to the right, while ∣x+h∣ is translated h units to the left.Finally, we have a vertical translation 5 units down, ∣-x+1∣⇒∣-x+1∣−5.Graphing j(x) To graph the function without going through the entire process of transforming the parent function, we can make a table of values. Then we only need to plot the ordered pairs that we find.x∣-x+1∣−5Simplifyj(x) -3∣-(-3)+1∣−5∣4∣−5-1 -1∣-(-1)+1∣−5∣2∣−5-3 1∣-(1)+1∣−5∣0∣−5-5 3∣-(3)+1∣−5∣-2∣−5-3 5∣-(5)+1∣−5∣-4∣−5-1 Now we can plot these points and connect them to create our graph of j(x).
Exercises 40 First we will go through all the transformations from the graph of f(x)=∣x∣ to the graph of given function. After that we will make a table of values to graph this function.Transformations We are given the following absolute value equation. n(x)=∣∣​-31​x+1∣∣​+2​ This is a transformation of the parent function f(x)=∣x∣. To see how each of the parameters is affecting the parent function it will be helpful to rewrite it first. n(x)=∣∣​-31​x+1∣∣​+2 ⇔ n(x)=∣∣​-31​(x−3)∣∣​+2​ This way we can look at each of the transformations individually and work our way from the "inside" to the "outside." We will define each transformation, starting with that which is closest to x and end with the farthest from x. First we have a horizontal translation 3 units to the right. ∣x∣⇒∣x−3∣.Then we have a horizontal stretch by a factor of 3, ∣x−3∣⇒∣∣​31​(x−3)∣∣​.Next we have a reflection in the y-axis, ∣∣​31​(x−3)∣∣​⇒∣∣​-31​(x−3)∣∣​. Notice that, since the graph is symmetrical, this will not change how it looks.Finally, we have a vertical translation 2 units up, ∣∣​-31​(x−3)∣∣​⇒∣∣​-31​(x−3)∣∣​+2.Graphing n(x) To graph the function without going through the entire process of transforming the parent function, we can make a table of values. Then we only need to plot the ordered pairs that we find.x∣∣∣∣​-31​x+1∣∣∣∣​+2Simplifyn(x) -3∣∣∣∣​-31​(-3)+1∣∣∣∣​+2∣2∣+24 0∣∣∣∣​-31​(0)+1∣∣∣∣​+2∣1∣+23 3∣∣∣∣​-31​(3)+1∣∣∣∣​+2∣0∣+22 6∣∣∣∣​-31​(6)+1∣∣∣∣​+2∣-1∣+23 Now we can plot these points and connect them to create our graph of n(x).
Exercises 41 aLet's make a table to graph the function. To do so, we will assign some values to the x-variable and obtain their corresponding y-values.t-2∣t−15∣+50s(t)=-2∣t−15∣+50 0-2∣0−15∣+5020 10-2∣10−15∣+5040 15-2∣15−15∣+5050 20-2∣20−15∣+5040 30-2∣30−15∣+5020 Let's now plot and connect the obtained points. Keep in mind that the graph of an absolute value function has a V shape. Also, since neither the weeks nor the number of shoes sold can be negative, we will draw our graph only in the first quadrant.bThe greatest number of pairs sold in one week is represented by the vertex of the function. Let's look at the graph and identify where this occurs.We see that the vertex is (15,50). In the context of our exercise, this means that in the 15th week, 50000 pairs of shoes were sold.
Exercises 42 aIn this exercise, we are asked to use a graph and equation of an absolute value function to determine the point where a shot results in hitting the side of a pool table. It is given that the side of the table corresponds with the x-axis. Thus, the ball will hit the side at the same point where the graph intersects the x-axis. In general, a function will intersect the x-axis when y=0. Thus, we can substitute y=0 into the given equation and solve for x. y=34​∣∣∣∣​x−45​∣∣∣∣​y=00=34​∣∣∣∣​x−45​∣∣∣∣​LHS⋅43​=RHS⋅43​0=∣∣∣∣​x−45​∣∣∣∣​Rearrange equation∣∣∣∣​x−45​∣∣∣∣​=0 Since the absolute value equals 0 we get only one possible solution when removing the absolute value x−45​=0⇔x=45​.​ The solution x=45​ means that the graph intersects the x-axis at (45​,0). Thus, the ball hits the wall at this point.bWe can assume that the shot is intended so that the ball hits the top left corner pocket. From the graph, this is the point (-5,5). If the shot is made, then (-5,5) will be a solution to the given equation. We will substitute -5 for x and 5 for y. y=34​∣∣∣∣​x−45​∣∣∣∣​x=-5, y=55=?34​∣∣∣∣​-5−45​∣∣∣∣​ Simplify RHS Write as a fraction5=?34​∣∣∣∣​-420​−45​∣∣∣∣​Subtract terms5=?34​∣∣∣∣​-425​∣∣∣∣​∣∣∣∣​-425​∣∣∣∣​=425​5=?34​⋅425​Multiply fractions5=?12100​Calculate quotient 5≠8.33 The point (-5,5) is not a solution to the equation. Thus, the shot is not made.
Exercises 43 aThe function g(x)=f(x)−a transforms f(x) by shifting it down a units. Thus, g(x)=f(x) − 5 vertically shifts f down by 5 units. To determine the points of g, we can decrease the y-coordinates of f's points by 5. The x-coordinates will remain the same, because no horizontal shift takes place. The following table gives the transformed points.Pointf(x)f(x)−5g(x) A(-21​,3)(-21​,3−5)(-21​,-2) B(1,0)(1,0−5)(1,-5) C(-4,-2)(-4,-2−5)(-4,-7)bThe function h(x)=f(x−a) transforms f(x) by shifting it right a units. Thus, h(x)=f(x − 3) horizontally shifts f right by 3 units. To determine the points of h, we can increase the x-coordinates of f's points by 3. The y-coordinates will remain the same, because no vertical shift takes place. The following table gives the transformed points.Pointf(x)f(x−3)h(x) A(-21​,3)(-21​+3,3)(221​,3) B(1,0)(1+3,0)(4,0) C(-4,-2)(-4+3,-2)(-1,-2)cThe function j(x)=-f(x) transforms f(x) by reflecting it over the x-axis. To determine the points of j, we can multiply the y-coordinates of f's points by -1. The x-coordinates will remain the same, because no horizontal shift takes place. The following table gives the transformed points.Pointf(x)-1f(x)j(x) A(-21​,3)(-21​,3⋅(-1))(-21​,-3) B(1,0)(1,0⋅(-1))(1,0) C(-4,-2)(-4,-2⋅(-1))(-4,2)dThe function k(x)=a⋅f(x) transforms f(x) vertically stretching it by a factor of a. Thus k(x)=4f(x) vertically stretches f by a factor of 4. To determine the points of k, we can multiply the y-coordinates of f's points by 4. The x-coordinates will remain the same, because no horizontal shift takes place. The following table gives the transformed points.Pointf(x)k(x)=4f(x)j(x) A(-21​,3)(-21​,3⋅4)(-21​,12) B(1,0)(1,0⋅4)(1,0) C(-4,-2)(-4,-2⋅4)(-4,-8)
Exercises 44 aThe transformation y=∣x∣+k vertically shifts the graph of y=∣x∣ by k units. We can arbitrarily choose values for k to determine the direction of the translation.If k>0, the graph of y=∣x∣+k is a translation up k units of the graph of y=∣x∣. For example, y=∣x∣+4 is a translation up 4 units of the graph of y=∣x∣. If k<0, the graph of y=∣x∣+k is a translation down k units of the graph of y=∣x∣. For example, y=∣x∣−4 is a translation down 4 units of the graph of y=∣x∣.Let's graph the examples we have stated along with y=∣x∣.bThe transformation y=∣x−h∣ horizontally shifts the graph of y=∣x∣ by h units. We can arbitrarily choose values for h to determine the direction of the translation.If h>0, the graph of y=∣x−h∣ is a translation right h units of the graph of y=∣x∣. For example, y=∣x−3∣ is a translation right 3 units of the graph of y=∣x∣. If h<0, the graph of y=∣x−h∣ is a translation left h units of the graph of y=∣x∣. For example, y=∣x+3∣ is a translation left 3 units of the graph of y=∣x∣.Let's graph the examples we have stated along with y=∣x∣.cThe transformation y=a∣x∣ vertically stretches y=∣x∣ by a factor of a. We can arbitrarily choose values for a to determine how the graph changes.If a>0, the graph of y=a∣x∣ is a vertical stretch of the graph of y=∣x∣ by a factor of a. For example, the graph of y=2∣x∣ is a vertical stretch of the graph of y=∣x∣ by a factor of 2. If a<0, the graph of y=a∣x∣ is a vertical stretch of the graph of y=∣x∣ by a factor of ∣a∣, followed by a reflection in the x-axis. For example, the graph of y=-2∣x∣ is a vertical stretch of the graph of y=∣x∣ by a factor of 2, followed by a reflection in the x-axis.Let's graph the examples we have stated along with y=∣x∣ to show both transformations.dThe transformation y=∣ax∣ horizontally shrinks the graph of y=∣x∣ by a factor of ∣a∣. Since the absolute value removes the negative sign, it does not matter whether a is a positive or a negative number. Let's show this by graphing y=∣x∣, y=∣3x∣, and y=∣-3x∣ on the same coordinate plane.
Exercises 45 To begin, y=∣x−h∣+k transforms the parent function y=∣x∣ by horizontally shifting it by h units and vertically translating it by k units. From the equation, y=∣x−1∣−3, we can see that h=1 and k=-3. Thus, y=∣x∣ is shifted to the right by 1 unit and down 3 units. The horizontal shift contains the mistake. Notice on the given graph that y=∣x∣ was shifted 1 unit to the left but when we have a subtraction from x the shift occurs to the right. We will graph the function correctly below.
Exercises 46 To begin, y=a∣x∣ vertically stretches y=∣x∣ by a factor of a. Additionally, if a is negative, the graph will be reflected across the x-axis. However, the function was not reflected over the x-axis now was it. We will graph the function correctly below.
Exercises 47 To begin creating our square, we need to determine the length we want for the two known sides. To do this, let's draw the graph of y=∣x∣−2 with the restricted domain -2≤x≤2.We can now reflect the above graph in the x-axis to create the missing sides of the square. This means that we need to multiply the outcome of the function by -1. gOriginal Functiong​y=∣x∣−2​​gReflected Functiong​y=-(∣x∣−2)  ⇔  y=-∣x∣+2​ The function whose graph will make a square with the graph of the given function is y=-∣x∣+2. We will add this to the graph and restrict the domain to -2≤x≤2.
Exercises 48 We want to write an absolute value function whose graph forms a square with the given graph. To do so, we will need to draw any horizontal line. Let's arbitrarily choose the line y=3.Note that, due to the symmetry of the absolute value function, we have an isosceles triangle. Since one of the angles is a right angle, the measure of the other two angles must be 45∘. Let's now translate our graph down 3 units. We obtain this transformation by subtracting 3 units from the output. y=∣x−3∣+1−3⇔y=∣x−3∣−2​ Now the triangle will be formed with the x-axis.Let's now reflect our graph in the x-axis. We obtain this by changing the sign of the output. y=-(∣x−3∣−2)⇔y=-∣x−3∣+2​ Let's graph the above function.Note that the triangles above are congruent. Therefore, the obtained triangle is an isosceles triangle with one right angle and two angles whose measure is 45∘. If we merge these isosceles triangles, we obtain a square.Finally, we have to go back to the original position. To do so, we will translate both graphs up 3 units by adding 3 to the outputs. y=∣x−3∣−2+3   ⇕   y=∣x−3∣+1   ​   y=-∣x−3∣+2+3   ⇕   y=-∣x−3∣+5​ Let's see the final graph.Note that there are infinitely many functions whose graphs form a square with the given graph. The function we found, y=-∣x−3∣+5, is just one of them.
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Exercises 50 aBefore we can figure out how to correct the path of the ball so that it will hit the brick, let's look at the uninterrupted path it will take.If we move our paddle out of the way, rather than using it to help us hit the brick, we will miss the target by 2 steps to the right. However, because the ball will always deflect at a right angle, we know that the slope will always change in the same way: Before deflection: After deflection: ​m=-1m=1.​ How can we use this to figure out where can we place our paddle so that the ball is deflected in just the right way to hit the brick? Let's trace our steps backwards from the brick and see where it intersects with the original path.If we place our paddle at (7,1), the ball will change directions in time to accurately hit the brick.bLet's compare the graph of the ball's path to the parent absolute value function. First, we need to graph them on the same coordinate plane.How can we write a transformation of the parent function such that the new function will be the same as the ball's path? The slopes are the same for both function so we know that it will only be translations. So, how can we get from the parent vertex to the transformed vertex?The graph is shifted by 1 unit up, a vertical translation, and 7 units to the right, a horizontal translation. Translations work in a way such that: Vertical translation: Horizontal translation: ​g(x)=f(x)+kg(x)=f(x−h),​ where k is the number of units the graph is shifted up (for positive numbers) or down (for negative numbers), and h is the number of units the graph is shifted right (for positive numbers) or left (for negative numbers). Our graph's equation then becomes: y=∣x−7∣+1.
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