#### Transformations of Graphs of Linear Functions

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Exercises 1 The function f(x)=x is considered a parent function because all other non-constant linear functions are transformations of it. The transformations can be translations, reflections, stretches, or shrinks, but all non-constant linear functions are just manipulations of the parent function.
Exercises 2 The four main types of transformations are: translating, reflecting, stretching, and shrinking. Let's look at each of these one by one.Translations The translation of a function can either be vertical or horizontal. Let's begin with looking at a fairly basic function, f(x)=2x.If we want to vertically translate this function up by 2, then we have to take f(x) and add 2. Our new function then becomes: y=f(x)+2⇒y=2x+2.If we want to horizontally translate this function by 2 to the right, we have y=f(x−2). Our function then becomes: y=2(x−2)⇒y=2x−4Reflections The reflection of a function can either be across either axis. Let's begin with looking at the function, f(x)=x+1.A reflection across the x-axis requires us to reverse the sign of each y-value, we have y=-f(x). Our function then becomes: y=-(x+1)⇒y=-x−1A reflection across the y-axis requires us to reverse the sign of each x-value, we have y=f(-x). Our function then becomes: y=-x+1Stretches The stretching of a function can either be vertical or horizontal. Let's begin with looking at the function f(x)=x+2.If we want to vertically stretch this function by a factor of 3, this means that we multiply each y-value by 3. We then have y=3⋅f(x) and our function becomes: y=3(x+2)⇒y=3x+6If we want to horizontally stretch this function by a factor of 4, this means that we multiply each x-value by 41​. Remember that, with horizontal stretches and shrinks, we need to multiply each x-value by a1​ because we want to stretch or shrink away from the y-axis. We then have y=f(41​x) and our function becomes: y=41​x+2Shrinks The shrinking of a function is very similar to the stretching of a function, it is the same basic process except the factor by which you shrink things is a non-negative fraction in the form of a1​. Similar to stretching, it can either be vertical or horizontal. Let's begin with looking at the same function as we did with our stretching, f(x)=x+2.If we want to vertically shrink this function by a factor of 31​, this means that we multiply each y-value by 31​. We then have y=31​⋅f(x) and our function becomes: y=31​(x+2)⇒y=31​x+32​If we want to horizontally shrink this function by a factor of 41​, this means that we multiply each x-value by 4. Remember that, with horizontal stretches and shrinks, we need to multiply each x-value by a1​ because we want to stretch or shrink away from the y-axis. a=41​⇒1÷41​=4 We then have y=f(4x) and our function becomes: y=4x+2
Exercises 3 If our base function is y=f(x), then multiplying either all of the x-values or all of the y-values by a positive number will affect the slope of the function. Let's look at each of the functions given.y=f(ax) The graph of y=f(ax) is a horizontal stretch or shrink of y=f(x). We know that it is horizontal because the x-values are changing, they are moving left and right, and the function is either stretching away from or shrinking towards the y-axis. It will be considered a stretch if 0<a<1, because each value of x produces a smaller value of y. The function is staying nearer to the x-axis and is being stretched in the left and right directions. It will be considered a shrink if a>1, because each value of x produces a larger value of y. The function is staying farther from the x-axis and is being shrunk with respect to the left and right directions.y=af(x) The graph of y=af(x) is a vertical stretch or shrink of y=f(x). We know that it is vertical because the y-values are changing, they are moving up and down, and the function is either stretching away from or shrinking towards the x-axis. It will be considered a stretch if a>1, because each value of y is becoming larger, or "stretched." It will be considered a shrink if 0<a<1, because each value of y is becoming smaller, or "shrunk."
Exercises 4 The x-intercepts in both functions appear to be the same. Let's check to see if this will always be the case. When we vertically stretch or shrink a function by any arbitrary factor, we multiply the entire function by that factor. Let's look at a generic equation, y=mx+b, and solve for the x-intercept. y=mx+by=00=mx+b Solve for x LHS−b=RHS−b-b=mxLHS/m=RHS/mm-b​=xRearrange equation x=-mb​ If we want to vertically shrink this function, we multiply f(x) by a1​. Let's do this and solve for our new x-intercept. y=a1​(mx+b)y=00=a1​(mx+b) Solve for x LHS⋅a=RHS⋅aa⋅0=mx+bZero Property of Multiplication0=mx+bLHS−b=RHS−b-b=mxLHS/m=RHS/mm-b​=xRearrange equation x=-mb​ Both times the x-intercept was x=-mb​. Therefore, a vertical shrink will never change the x-intercept.
Exercises 5 We are given the graphs of f and g on a coordinate plane, and want to describe the transformation from the graph of f to the graph of g.We see that the graph of g is a vertical translation 2 units up of the graph of f. In fact, the function g has the form y=f(x)+k. g(x)=f(x)+2​ In our function, we have that k=2. This means the graph of g is a horizontal translation 2 units up of the graph of f.
Exercises 6 We are given the graphs of f and g on a coordinate plane, and want to describe the transformation from the graph of f to the graph of g.We see that the graph of g is a horizontal translation 4 units left of the graph of f. In fact, the function g has the form y=f(x−h). g(x)=f(x+4)⇔g(x)=f(x−(-4))​ In our function, we have that h=-4. This means the graph of g is a horizontal translation 4 units left of the graph of f.
Exercises 7 First, let's use a table of values to find points on the graph of f(x)=31​x+3x31​x+3f(x) -331​(-3)+32 031​(0)+33 331​(3)+34 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function g(x)=f(x)−3 differs from f(x).xf(x)f(x)−3g(x) -322−3-1 033−30 344−31 If we plot these points on the same coordinate plane as f(x), we can compare the two functions.We see that each y-value is being translated 3 units down. So, the graph of g is a vertical translation 3 units down of the graph f.
Exercises 8 First, let's use a table of values to find points on the graph of f(x)=-3x+4.x-3x+4f(x) -2-3(-2)+410 0-3(0)+44 2-3(2)+4-2 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function g(x)=f(x)+1 differs from f(x).xf(x)f(x)+1g(x) -21010+111 044+15 2-2-2+1-1 If we plot these points on the same coordinate plane as f(x), we can compare the two functions.We see that each y-value is being translated 1 units up. So, the graph of g is a vertical translation 1 unit up of the graph of f.
Exercises 9 First, let's use a table of values to find points on the graph of f(x)=-x−2.x-x−2f(x) -2-(-2)−20 0-(0)−2-2 2-(2)−2-4 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function g(x)=f(x+5) differs from f(x).xf(x)f(x+5)g(x) -75-(-7+5)−20 -53-(-5+5)−2-2 -31-(-3+5)−2-4 If we plot these points on the same coordinate plane as f(x), we can compare the two functions.We see that each y-value is being translated 5 units left. So, the graph of g is a horizontal translation 5 units left of the graph of f.
Exercises 10 First, let's use a table of values to find points on the graph of f(x)=21​x−5.x21​x−5f(x) -221​(-2)−5-6 021​(0)−5-5 221​(2)−5-4 We can plot these points and connect them with a straight line to have the graph of f(x).Function g(x) Now, let's look at how the function g(x)=f(x−3) differs from f(x).xf(x)f(x−3)g(x) 1-421​21​(1−3)−5-6 3-321​21​(3−3)−5-5 5-221​21​(5−3)−5-4 If we plot these points on the same coordinate plane as f(x), we can compare the two functions.We see that each y-value is being translated 3 units to the right. So the graph of g is a horizontal translation 3 units right of the graph of f.
Exercises 11 This exercise is actually very simple, we have been given a lot of extra information that we don't need in order to answer the question at hand. Ultimately, we only need to know what translation is being performed between the functions d(t) and f(t). d(t)→?f(t)=d(t−5) We can eliminate a few options (reflections, stretches, and shrinks) because any time we are adding to or subtracting from an equation, it is a translation. So, is this a vertical translation or a horizontal translation? A vertical translation means that the output is altered after the equation has already been simplified. The general form for this type of transformation is: g(x)=f(x)+k. A horizontal translation means that the input is altered before the equation has been simplified. The general form for this type of transformation is: h(x)=f(x−h). The given equation f(t)=d(t−5) is changing the values of t before they are being inputted into the function d(t). This means that the transformation is a horizontal translation by 5 units in the positive direction.
Exercises 12 This exercise is actually very simple, we have been given a lot of extra information that we don't need in order to answer the question at hand. Ultimately, we only need to know what translation is being performed between the functions C(p) and T(p). C(p)→?T(p)=C(p)+25. We can eliminate a few options (reflections, stretches, and shrinks) because any time we are adding to or subtracting from an equation, it is a translation. Is this a vertical translation or a horizontal translation? A vertical translation means that the output is altered after the equation has already been simplified. The general form for this type of transformation is: g(x)=f(x)+k. A horizontal translation means that the input is altered before the equation has been simplified. The general form for this type of transformation is: h(x)=f(x−h). The given equation T(p)=C(p)+25 is changing the values of C(p) after the original function has been completed. This means that the transformation is a vertical translation by 25 units in the positive direction.
Exercises 13 We are given the graphs of f and h on a coordinate plane, and want to describe the transformation from the graph of f to the graph of h. We see that the points on the graphs of f and h have the same x-coordinate, but opposite y-coordinates.The above means that the x-axis acts as a line of symmetry. Therefore, the graph of h is a reflection of the graph of f in the x-axis. Furthermore, note that h has the form y=-f(x). This means, as we have already said, that the graph of h is a reflection in the x-axis of the graph of f.
Exercises 14 We are given the graphs of f and h on a coordinate plane, and want to describe the transformation from the graph of f to the graph of h. We see that the points on the graphs of f and h have the same y-coordinate, but opposite x-coordinates.The above means that the y-axis acts as a line of symmetry. Therefore, the graph of h is a reflection in the y-axis of the graph of f. Furthermore, note that h has the form y=f(-x). This means, as we have already said, that the graph of h is a reflection in the y-axis of the graph of f.
Exercises 15 First, let's use a table of values to find points on the graph of f(x)=-5−x.x-5−xf(x) -2-5−(-2)-3 0-5−(0)-5 2-5−(2)-7 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function h(x)=-f(x) differs from f(x).xf(x)f(-x)h(x) -2-3-5−(−(-2))-7 0-5-5−(−(0))-5 2-7-5−(−(2))-3 If we plot these points on the same coordinate plane as f(x), we can compare the two functions.We see that each y-value is flipped over the y-axis. So the graph of h is a reflection in the y-axis of the graph of f.
Exercises 16 First, let's use a table of values to find points on the graph of f(x)=41​x−2.x41​x−2f(x) -441​(-4)−2-3 041​(0)−2-2 441​(4)−2-1 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function h(x)=-f(x) differs from f(x).xf(x)-f(x)h(x) -4-3-(-3)3 0-2-(-2)2 4-1-(-1)1 If we plot these points on the same coordinate plane as f(x), we can compare the two functions.We see that each x-value is flipped over the x-axis. So the graph of h is a reflection in the x-axis of the graph of f.
Exercises 17 The graph of y=a⋅f(x) is a vertical stretch or shrink by a factor of a of the graph of y=f(x) as long as two conditions are met. a>0  and  a≠1​ A vertical stretch means the new function is going further away from the x-axis and a vertical shrink means it is coming closer to the x-axis. Whether the transformation is a stretch or a shrink depends on the value of a. Vertical stretch: Vertical shrink: ​a>10<a<1​ From the graph, we see that the function r(x)=2f(x) is going further away from the x-axis. Hence, the graph of r is a vertical stretch of the graph of f by a factor of 2.
Exercises 18 The graph of y=f(ax) is a horizontal stretch or shrink by a factor of a1​ of the graph of y=f(x) as long as two conditions are met. a>0 and a≠1​ A horizontal stretch means the new function is going further away from the y-axis and a horizontal shrink means it comes closer to the y-axis. Whether the transformation is a stretch or shrink depends on the value of a. Horizontal shrink: Horizontal stretch: ​a>10<a<1​ From the graph, we see that the function r(x)=f(41​x) is going further away from the y-axis. In other words, the graph of r is a horizontal stretch of the graph of f by a factor of 4.
Exercises 19 First, let's use a table of values to find points graph of f(x)=-2x−4.x-2x−4f(x) 0-2(0)−4-4 2-2(2)−4-8 4-2(4)−4-12 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function r(x)=f(21​x) differs from f(x).xf(x)f(21​x)r(x) 0-4-2(21​(0))−4-4 2-8-2(21​(2))−4-6 4-12-2(21​(4))−4-8 If we plot these points on the same coordinate plane as f(x), we can see that our points have been stretched twice as far from the y-axis as they were before.Therefore the graph of r is a horizontal stretch of the graph of f by a factor of 2.
Exercises 20 First, let's use a table of values to find points graph of f(x)=3x+5.x3x+5f(x) 03(0)+55 23(2)+511 43(4)+517 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function r(x)=f(31​x) differs from f(x).xf(x)f(31​x)r(x) 053(31​(0))+55 3143(31​(3))+58 6233(31​(6))+511 If we plot these points on the same coordinate plane as f(x), we can see that our points have been stretched three times as far from the y-axis as they were before.Therefore the graph of r is a horizontal stretch of the graph of f by a factor of 3.
Exercises 21 First, let's use a table of values to find points graph of f(x)=32​x+1.x32​x+1f(x) 032​(0)+11 332​(3)+13 632​(6)+15 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function r(x)=2f(x) differs from f(x).xf(x)2f(x)r(x) 013(1)3 333(3)9 653(5)15 If we plot these points on the same coordinate plane as f(x), we can see that our points have been stretched three times as far from the x-axis as they were before.Therefore the graph of r is a vertical stretch of the graph of f by a factor of 3.
Exercises 22 First, let's use a table of values to find points graph of f(x)=-41​x−2.x-41​x−2f(x) 0-41​(0)−2-2 4-41​(4)−2-3 8-41​(8)−2-4 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function r(x)=4f(x) differs from f(x).xf(x)4f(x)r(x) 0-24(-2)-8 4-34(-3)-12 8-44(-4)-16 If we plot these points on the same coordinate plane as f(x), we can see that our points have been stretched four times as far from the x-axis as they were before.Therefore the graph of r is a vertical stretch of the graph of f by a factor of 4.
Exercises 23 The graph of y=f(ax) is a horizontal stretch or shrink by a factor of a1​ of the graph of y=f(x) as long as two conditions are met. a>0 and a≠1​ A horizontal stretch means the new function is going further away from the y-axis and a horizontal shrink means it comes closer to the y-axis. Whether the transformation is a stretch or shrink depends on the value of a. Horizontal shrink: Horizontal stretch: ​a>10<a<1​From the graph, we see that the function h(x)=f(3x) is coming closer to the y-axis. In other words, the graph of h is a horizontal shrink of the graph of f by a factor of 31​.
Exercises 24 The graph of y=a⋅f(x) is a vertical stretch or shrink by a factor of a of the graph of y=f(x) as long as two conditions are met. a>0  and  a≠1​ A vertical stretch means the new function is going further away from the x-axis and a vertical shrink means it is coming closer to the x-axis. Whether the transformation is a stretch or a shrink depends on the value of a. Vertical stretch: Vertical shrink: ​a>10<a<1​From the graph, we see that the function h(x)=31​f(x) is coming closer to the x-axis. Hence, the graph of h is a vertical shrink of the graph of f by a factor of 31​.
Exercises 25 First, let's use a table of values to find points graph of f(x)=3x−12.x3x−12f(x) 03(0)−12-12 23(2)−12-6 43(4)−120 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function h(x)=61​f(x) differs from f(x).xf(x)61​f(x)h(x) 0-1261​(-12)-2 2-661​(-6)-1 4061​(0)0 If we plot these points on the same coordinate plane as f(x), we can see that our points have been shrunk six times as close to the x-axis as they were before.Therefore, the graph of h is a vertical shrink of the graph of f by a factor of 61​.
Exercises 26 First, let's use a table of values to find points graph of f(x)=-x+1.x-x+1f(x) 0-(0)+11 2-(2)+1-1 4-(4)+1-3 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function h(x)=f(2x) differs from f(x).xf(x)f(2x)h(x) 01-(2(0))+11 2-1-(2(2))+1-3 4-3-(2(4))+1-7 If we plot these points on the same coordinate plane as f(x), we can see that our points have been shrunk to half as far away from the y-axis as they were before.Therefore the graph of h is a horizontal shrink of the graph of f by a factor of 21​.
Exercises 27 First, let's use a table of values to find points graph of f(x)=-2x−2.x-2x−2f(x) 0-2(0)−2-2 5-2(5)−2-12 10-2(10)−2-22 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function h(x)=f(5x) differs from f(x).xf(x)f(5x)h(x) 0-2-2(5(0))−2-2 1-4-2(5(1))−2-12 2-6-2(5(2))−2-22 If we plot these points on the same coordinate plane as f(x), we can see that our points have been shrunk to be one fifth as far away from the y-axis as they were before.Therefore, the graph of h is a horizontal shrink of the graph of f by a factor of 51​.
Exercises 28 First, let's use a table of values to find points graph of f(x)=4x+8.x4x+8f(x) 04(0)+88 24(2)+816 44(4)+824 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function h(x)=43​f(x) differs from f(x).xf(x)43​f(x)h(x) 0843​(8)6 21643​(16)12 42443​(24)18 If we plot these points on the same coordinate plane as f(x), we can see that our points have been shrunk to three-quarters as far away from the x-axis as they were before.Therefore, the graph of h is a vertical shrink of the graph of f by a factor of 43​.
Exercises 29 First, let's use a table of values to find points graph of f(x)=x−2.xx−2f(x) 0(0)−2-2 3(3)−21 6(6)−24 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function g(x)=41​f(x) differs from f(x).xf(x)41​f(x)g(x) 0-241​(-2)-21​ 3141​(1)41​ 6441​(4)1 If we plot these points on the same coordinate plane as f(x), we can see that our points have been shrunk to quarter as far away from the x-axis as they were before.Therefore the graph of g is a vertical stretch of the graph of f by a factor of 41​.
Exercises 30 First, let's use a table of values to find points graph of f(x)=-4x+8.x-4x+8f(x) 0-4(0)+88 2-4(2)+80 4-4(4)+8-8 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function g(x)=-f(x) differs from f(x).xf(x)-f(x)g(x) 08-(8)-8 20-(0)0 4-8-(-8)8 If we plot these points on the same coordinate plane as f(x), we can compare the two functions.We see that each point is flipped over the x-axis. Therefore the graph of g is a reflection in the x-axis of the graph of f.
Exercises 31 First, let's use a table of values to find points on the graph of f(x)=-2x−7.x-2x−7f(x) -2-2(-2)−7-3 0-2(0)−7-7 2-2(2)−7-11 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function g(x)=f(x−2) differs from f(x).xf(x)f(x−2)g(x) -2-3-2(-2−2)−71 0-7-2(0−2)−7-3 2-11-2(2−2)−7-7 If we plot these points on the same coordinate plane as f(x), we can compare the two functions.We see that each y-value is being translated 2 units right. Therefore the graph of g is horizontal translation 2 units right of the graph of f.
Exercises 32 First, let's use a table of values to find points graph of f(x)=3x+8.x3x+8f(x) 03(0)+88 23(2)+814 43(4)+820 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function g(x)=f(32​x) differs from f(x).xf(x)f(32​x)g(x) 083(32​(0))+88 3173(32​(2))+814 6263(32​(4))+820 If we plot these points on the same coordinate plane as f(x), we can see that our points have been stretched one and a half times as far from the y-axis as they were before.Therefore, the graph of g is a horizontal stretch by a factor of 23​ of the graph of f.
Exercises 33 First, let's use a table of values to find points graph of f(x)=x−6.xx−6f(x) 0(0)−6-6 2(2)−6-4 4(4)−6-2 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function g(x)=6f(x) differs from f(x).xf(x)6f(x)g(x) 0-66(-6)-36 2-46(-4)-24 4-26(-2)-12 If we plot these points on the same coordinate plane as f(x), we can see that our points have been stretched to six times as far from the x-axis as they were before.Therefore the graph of g is a vertical stretch of the graph of f by a factor of 6.
Exercises 34 First, let's use a table of values to find points on the graph of f(x)=-x.x-xf(x) -2-(-2)2 0-(0)0 2-(2)-2 We can plot these points and connect them with a straight line to have the graph of f(x).Now, let's look at how the function g(x)=f(x)−3 differs from f(x).xf(x)f(x)−3g(x) -222−3-1 000−3-3 2-2-2−3-5 If we plot these points on the same coordinate plane as f(x), we can compare the two functions.We see that each y-value is being translated 3 units down. Therefore, the graph of g is a vertical translation 3 units down of the graph of f.
Exercises 35 A horizontal translation is a shift of a graph along the x-axis. This means we have to manipulate the inputs (the x-values) of the original function. To move a function f(x) by h units to the right, we have to subtract h from the function's inputs. We get: f(x−h). We have that "the graph of g should be a horizontal translation 2 units to the right of the graph of f," This means we have to subtract 2 from x: g(x)=f(x−2).
Exercises 36 A reflection is a transformation that flips a graph over a line called the line of reflection. A reflection in the y-axis, means that the y-values stay the same, but the signs of the x-values are changed. This means we have to multiply the inputs by -1 which changes the sign: g(x)=f(-x).
Exercises 37 A vertical stretch of a function f(x) by a factor 4, means that each output (y-value) is multiplied by 4. Therefore, to vertically stretch f(x) by 4, we can write: g(x)=4⋅f(x).
Exercises 38 A horizontal shrink by a factor of 51​ means that each x-value of the graph of f(x) is multiplied by 5. This allows matching y-values in g(x) to be 51​ of the distance away from the y-axis as they were in f(x). Then we have: g(x)=f(5x). This concept is not easy to understand in words so let's look at a graph of two functions. Here we have f(x)=x and g(x)=f(5x).
Exercises 39 The function g has the form y=f(x−h). f(x−2)​ We have here that h=2. Therefore, the graph of g is a horizontal translation 2 units right of the graph of f. In the graph shown in the exercise, the translation is done to the left. This is the error. Let's see how the correct graph looks like.
Exercises 40 We see that g is of the form y=f(-x). Let's make a table of values to graph g. To find its outputs, we will multiply the inputs by -1 and then evaluate f.x-xf(-x)g(x)=f(-x) -22-2+31 00-0+33 2-2-(-2)+35 Let's now plot these points and connect them with a straight line.We see that the graph of g is a reflection of the graph of f in the y -axis. In the graph shown in the exercise, the reflection is done in the x-axis. This is the error.
Exercises 41 Let's begin by graphing both functions. Then, we can compare the differences between them.First, look at how the slope changes between f(x) and h(x).Looking at the slope triangles, we can see that the rise in f(x) is 3 times greater than the rise of h(x). This means that h(x) is moving away from the x-axis 31​ times as quickly as f(x). slope of f(x): slope of h(x): ​1 ⇔runrise​=66​31​⇔runrise​=62​​ When a function is pulled away from or pushed towards the x-axis, this is either a vertical stretch or shrink. Because the transformed function is closer to the x-axis, it is a vertical shrink by a factor of 31​. The next thing to notice is how the y-intercept changes.The y-intercept moved up 1 unit. This is a vertical translation up by 1.
Exercises 42 Let's begin by graphing both functions. Then we can compare them.First, look at how the slope changes between f(x) and h(x).Looking at the slopes we can see that, for a run of 3, the rise of h is 12, and the rise of f is 3. This means the slope of h is 4 times greater than the slope of f. slope of  f(x): slope of  h(x): ​runrise​=33​ =1runrise​=312​=4​ The above indicates that we have a vertical stretch by a factor of 4. The next thing to notice is how the y-intercept changes.The y-intercept is translated 2 units down. Therefore, the graph of h is a vertical stretch by a factor of 4, followed by a vertical translation 2 units down of the graph of f.
Exercises 43 How does the equation of h(x) differ from f(x)? Let's see! f(x)h(x)​=x=-3x − 4​ We can see that x was multiplied by -3. Thus, it was first multiplied by -1 and transformed to h1​(x)=-x. Then, h1​(x) was multiplied by 3 and transformed to h2​(x)=-3x. Let's see it on a graph!Finally, h2​(x) was translated 4 units down. Let's take a look at the final product!The graph of f was reflected across the x-axis, vertically stretched by a factor of 3, and translated 4 units down.
Exercises 44 How does the equation of h(x) differ from f(x)? Let's see! f(x)h(x)​=x=-21​x + 3​ We can see that x was multiplied by -21​. Thus, it was first multiplied by -1 and transformed to h1​(x)=-x. Then, h1​(x) was multiplied by 21​ and transformed to h2​(x)=-21​x. Let's see it on a graph!Finally, h2​(x) was translated 3 units up. Let's take a look at the final product!The graph of the parent function was reflected across the x-axis, vertically shrunk by a factor of 21​, and translated 3 units up.
Exercises 45 Let's begin by graphing both functions. Then we can compare the differences between them.First, look at how the slope changes between f(x) and h(x).Looking at the slope triangles, we can see that the rise in f(x) is 3 times smaller than the rise of h(x). This means that h(x) is moving away from the x-axis 3 times faster than f(x). slope of f(x): slope of h(x): ​2 ⇔runrise​=36​6⇔runrise​=318​​ When a function is pulled away from or pushed towards the x-axis, this is either a vertical stretch or shrink. Because the transformed function is further away from the x-axis, it is a vertical stretch by a factor of 3. The next thing to notice is how the y-intercept changes.The y-intercept moved down 5 units. This is a vertical translation down by 5.
Exercises 46 How does the equation of h(x) differ from f(x)? Let's see! f(x)h(x)​=-3x=-3x − 7​ We can see that x was multiplied by -1. Thus, it was first multiplied by -1 and transformed to h1​(x)=-3x. Let's see it on a graph!Then, h1​(x) was translated 7 units down. Let's take a look at the final product!The graph of the parent function was reflected across the x-axis and translated 7 units down.
Exercises 47 Consider the given functions. d(x)=4x+72andt(x)=-4x+72​ We see that the slope of d is 4, and the slope of t is -4. Moreover, the y-intercept of both lines is 72. Let's use this information to graph the lines.As we have already said, the y-intercept is the same for both lines. Therefore, neither a translation nor a reflection in the x-axis are possible, since these transformations would change the position of the y-intercept. Is it possible a reflection in the y-axis? Let's see …We see that the transformation from the graph of t to the graph of d is a reflection in the y-axis. In fact, we can also identify this transformation by considering the equations. t(x)=-4x+72x=-xt(-x)=-4(-x)+72-a(-b)=a⋅bt(-x)=4x+27 Note that the expression on the right-hand side of the obtained formula is the function rule for d(x). t(-x)=4x+27⇔t(-x)=d(x)​ The above relation between the equations determines a reflection in the y-axis.
Exercises 48 Consider the given functions. P(x)=8x−150andQ(x)=16x−200​ Note that the slope and y-intercept of P are 8 and -150, respectively. Similarly, the slope and the y-intercept of Q are 16 and -200, respectively. Let's use this information to draw the lines.The slope of Q is twice the slope of P. Let's multiply the function rule of P by 2. This determines a vertical stretch by a factor of 2. 2P(x)P(x)=8x−1502(8x−150)Distribute 216x−300 To obtain the function rule for Q from the above expression, we need to add 100. This determines a vertical translation 100 units up. 16x−300+100=16x−200​ The right-hand side of the above equation is the function rule for Q. We can now describe the transformations from the graph of P to the graph of Q. Vertical stretch by a factor of 2, followedby vertical translation 100 units up.​
Exercises 49 aBy substituting a=4, b=0 and c=0 into the equation of g(x), we can determine how the graph of f was transformed. g(x)=a⋅f(x−b)+cSubstitute valuesg(x)=4⋅f(x−0)+0Simplify termsg(x)=4f(x) All the y-values are multiplied by 4, a value greater than 1. Therefore, this transformation is a vertical stretch of f.bBy substituting a=1, b=2 and c=0 into the equation of g(x), we can determine how the graph of f was transformed. g(x)=a⋅f(x−b)+cSubstitute valuesg(x)=1⋅f(x−2)+0Simplify termsg(x)=f(x−2) The graph of g(x) is a horizontal translation 2 units right of the graph of f.cBy substituting a=1 and b=0 into the equation of g(x), we can determine the value of c. g(x)=a⋅f(x−b)+ca=1, b=0g(x)=1⋅f(x−0)+cSimplify termsg(x)=f(x)+c Since we are adding c to the function rule, this represents a vertical translation of the graph of f. As we want a vertical translation 1 unit up, the value of c must be 1.
Exercises 50 aBy substituting a=31​, b=1, c=0, and d=0 into the equation of h(x), we can determine how the graph of f was transformed. h(x)=a⋅f(bx−c)+dSubstitute valuesh(x)=31​⋅f(1x−0)+0Simplify termsh(x)=31​f(x) All the y-values are multiplied by 31​, which is a smaller than 1. Therefore, this transformation is a vertical shrink of f.bBy substituting a=1, b=-1, c=0, and d=0 into the equation of h(x), we can determine how the graph of f was transformed. h(x)=a⋅f(bx−c)+dSubstitute valuesh(x)=1⋅f(-1x−0)+0Simplify termsh(x)=f(-x) The inputs of the function have their signs changed prior to being evaluated, this means h(x) is a reflection of f(x) in the y-axis.cBy substituting a=1, c=0, and d=0 into the equation of h(x), we can determine the value of b. h(x)=a⋅f(bx−c)+dSubstitute valuesh(x)=1⋅f(bx−0)+0Simplify termsh(x)=f(bx) This shows a horizontal stretch or shrink of the graph. Since we want this transformation to be a stretch by a factor of 5, the value of b must be 51​.
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