Chapter Test

Download for free
Find the solutions in the app
Android iOS
Exercises marked with requires Mathleaks premium to view it's solution in the app. Download Mathleaks app on Google Play or iTunes AppStore.
Sections
Exercises
Exercise name Free?
Exercises 1 Let's start by recalling the vertex form of a quadratic function. f(x)=a(x−h)2+k​ In this form the vertex of the parabola is the point (h,k), and the axis of symmetry is the vertical line x=h. Now consider the given function. h(x)=2x2−3⇕h(x)=2(x−0)2+(-3)​ We can see that h=0 and that k=-3. Therefore, the vertex is (0,-3), and the axis of symmetry is x=0. To graph the function we will make a table of values. Make sure to include x-values to the left and to the right of the axis of symmetry.x2x2−3h(x)=2x2−3 -22(-2)2−35 -12(-1)2−3-1 12(1)2−3-1 22(2)2−35 Let's now draw the parabola that connects the obtained points and the vertex. We will also draw the axis of symmetry x=0, and the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. Both graphs have the same axis of symmetry x=0. The graph of the given function is narrower than the graph of the parent function. The vertex of the given function, (0,-3), is below the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of h is a vertical stretch by a factor of 2 and a vertical translation down 3 units of the graph of f.
Exercises 2 Let's start by recalling the vertex form of a quadratic function. f(x)=a(x−h)2+k​ In this form the vertex of the parabola is the point (h,k), and the axis of symmetry is the vertical line x=h. Now consider the given function. g(x)=-21​x2⇕g(x)=-21​(x−0)2+0​ We can see that h=0 and that k=0. Therefore, the vertex is (0,0), and the axis of symmetry is x=0. To graph the function we will make a table of values. Make sure to include x-values to the left and to the right of the axis of symmetry.x-21​x2g(x)=-21​x2 -4-21​(-4)2-8 -2-21​(-2)2-2 2-21​(2)2-2 4-21​(4)2-8 Let's now draw the parabola that connects the obtained points and the vertex. We will also draw the axis of symmetry x=0, and the parent function f(x)=x2.From the graph above, we can note the following.The graph of the given function opens down, and the graph of the parent function opens up. Both graphs have the same axis of symmetry x=0. The graph of the given function is wider than the graph of the parent function. Both graphs have the same vertex, (0,0). From the graph and the observations above, we can conclude that the graph of g is a vertical shrink by a factor of 21​, and a reflection in the x-axis of the graph of f.
Exercises 3 Let's start by recalling the vertex form of a quadratic function. f(x)=a(x−h)2+k​ In this form the vertex of the parabola is the point (h,k), and the axis of symmetry is the vertical line x=h. Now consider the given function. p(x)=21​(x+1)2−1⇕p(x)=21​(x−(-1))2+(-1)​ We can see that h=-1 and that k=-1. Therefore, the vertex is (-1,-1), and the axis of symmetry is x=-1. To graph the function we will make a table of values. Make sure to include x-values to the left and to the right of the axis of symmetry.x21​(x+1)2−1p(x)=21​(x+1)2−1 -321​(-3+1)2−11 -221​(-2+1)2−1-21​ 021​(0+1)2−1-21​ 121​(1+1)2−11 Let's now draw the parabola that connects the obtained points and the vertex. We will also draw the axis of symmetry x=-1, and the parent function f(x)=x2.From the graph above, we can note the following.Both graphs open up. The graphs do not have the same axis of symmetry. The axis of symmetry of the parent function is the y-axis. The axis of symmetry of the given function is x=-1. The graph of the given function is wider than the graph of the parent function. The vertex of the given function, (-1,-1), is below and to the left of the vertex of the parent function, (0,0). From the graph and the observations above, we can conclude that the graph of p is a vertical shrink by a factor of 21​, a vertical translation down 1 unit, and a horizontal translation left 1 unit of the graph of f.
Exercises 4
Exercises 5 To draw the graph of the given function, we will follow five steps.Rewrite the quadratic function in intercept form. Identify and plot the x-intercepts. Find and graph the axis of symmetry. Identify the y-intercept and its reflection. Draw the parabola through the points we found.Let's go through these steps one at a time.Rewrite the Function We will start by rewriting the function in intercept form. To do so, we will factor the right-hand side of the given equation. f(x)=2x2−8x+8Factor out 2f(x)=2(x2−4x+4) Factor Write as a powerf(x)=2(x2−4x+22)Split into factorsf(x)=2(x2−2x(2)+22)a2−2ab+b2=(a−b)2f(x)=2(x−2)2Split into factors f(x)=2(x−2)(x−2)Identify and Plot the x-intercepts Recall the intercept form of a quadratic function. f(x)=a(x−p)(x−q)​ In this form, where a ​= 0, the x-intercepts are p and q. Let's consider the intercept form of our function. f(x)=2(x−2)2⇔f(x)=2(x−2)(x−2)​ We can see that a=2, p=2, and q=2. Therefore, the point (2,0) is the only one x-intercept. It means that this point is also the vertex of the given function.Find and Graph the Axis of Symmetry Since the parabola has only one x-intercept, which occurs at the point (2,0), the axis of symmetry is the vertical line x=2.Identifying the y-intercept and its Reflection We have found that in the given function the x-intercept is also the vertex, so we need to find other points to draw the parabola. We need at least three points. Recall the standard form in which the function was given. f(x)=2x2−8x+8​ The y-intercept of the graph of a quadratic function written in standard form is given by the value of c. Thus, the point where our graph intercepts the y-axis is (0,8). Let's plot this point and its reflection across the axis of symmetry.Draw the Parabola Finally, we will draw the parabola through the three points we have.We can see above that there are no restrictions on the x-variable. Furthermore, the y-variable takes values greater than or equal to 0. We can write the domain and range of the function using this information. Domain:Range:​ All real numbers y≥0​
Exercises 6 To draw the graph of the given function, we will follow four steps.Identify and plot the x-intercepts. Find and graph the axis of symmetry. Find and plot the vertex. Draw the parabola through the vertex and the points where the x-intercepts occur.Let's go through these steps one at a time.Identify and Plot the x-intercepts Recall the intercept form of a quadratic function. y=a(x−p)(x−q)​ In this form, where a ​= 0, the x-intercepts are p and q. Let's consider the intercept form of our function. y=-(x+5)(x−1)⇕y=-1(x−(-5))(x−1)​ We can see that a=-1, p=-5, and q=1. Therefore, the x-intercepts occur at (-5,0) and (1,0).Find and Graph the Axis of Symmetry The axis of symmetry is halfway between (p,0) and (q,0). Since we know that p=-5 and q=1, the axis of symmetry of our parabola is halfway between (-5,0) and (1,0). x=2p+q​⇒x=2-5+1​=2-4​=-2​ We found that the axis of symmetry is the vertical line x=-2.Find and Plot the Vertex Since the vertex lies on the axis of symmetry, its x-coordinate is -2. To find the y-coordinate, we will substitute -2 for x in the given equation. y=-(x+5)(x−1)x=-2y=-(-2+5)(-2−1) Simplify right-hand side Add and subtract termsy=-(3)(-3)a(-b)=-a⋅by=-(-9)-(-a)=a y=9 The y-coordinate of the vertex is 9. Therefore, the vertex is the point (-2,9).Draw the Parabola Finally, we will draw the parabola through the vertex and the x-intercepts.We can see above that there are no restrictions on the x-variable. Furthermore, the y-variable takes values less than or equal to 9. We can write the domain and range of the function using this information. Domain:Range:​ All real numbers y≤9​
Exercises 7 To draw the graph of the given function, we will follow five steps.Rewrite the quadratic function in intercept form. Identify and plot the x-intercepts. Find and graph the axis of symmetry. Find and plot the vertex. Draw the parabola through the vertex and the points where the x-intercepts occur.Let's go through these steps one at a time.Rewrite the Function We will start by rewriting the function in intercept form. To do so, we will factor the right-hand side of the given equation. h(x)=16x2−4Factor out 16h(x)=16(x2−41​)Write as a powerh(x)=16(x2−(21​)2)a2−b2=(a+b)(a−b)h(x)=16(x+21​)(x−21​)Identify and Plot the x-intercepts Recall the intercept form of a quadratic function. f(x)=a(x−p)(x−q)​ In this form, where a ​= 0, the x-intercepts are p and q. Let's consider the intercept form of our function. h(x)=16(x+21​)(x−21​)⇕h(x)=16(x−(-21​))(x−21​)​ We can see that a=16, p=-21​, and q=21​. Therefore, the x-intercepts occur at (-21​,0) and (21​,0).Find and Graph the Axis of Symmetry The axis of symmetry is halfway between (p,0) and (q,0). Since we know that p=-21​ and q=21​, the axis of symmetry of our parabola is halfway between (-21​,0) and (21​,0). x=2p+q​⇒x=2-21​+21​​=20​=0​ We found that the axis of symmetry is the vertical line x=0.Find and Plot the Vertex Since the vertex lies on the axis of symmetry, its x-coordinate is 0. To find the y-coordinate, we will substitute 0 for x in the given equation. h(x)=16x2−4x=0h(0)=16(0)2−4 Simplify right-hand side Calculate powerh(0)=16(0)−4Multiplyh(0)=0−4Subtract term h(0)=-4 The y-coordinate of the vertex is -4. Therefore, the vertex is the point (0,-4).Draw the Parabola Finally, we will draw the parabola through the vertex and the x-intercepts.We can see above that there are no restrictions on the x-variable. Furthermore, the y-variable takes values greater than or equal to -4. We can write the domain and range of the function using this information. Domain:Range:​ All real numbers y≥-4​
Exercises 8 We want to tell whether the table of values represents a linear, exponential, or quadratic function. To do so, we will analyze how the consecutive terms are related to each other.Note that the difference between consecutive x-values is always 1 — it is constant. Additionally, the consecutive y-values have a common ratio of 2. This means that the table represents an exponential function. Let's recall the general form of this type of function. y=abx​ We will use two ordered pairs given in the table to find the values of a and b. For simplicity, let's use (0,8) and (1,16). We will start by substituting 0 and 8 for x and y, respectively. y=abxx=0, y=88=ab0 Solve for a a0=18=a(1)Identity Property of Multiplication8=aRearrange equation a=8 We can write a partial equation of the function represented by the table. y=8bx​ To find the value of b we will substitute 1 for x and 16 for y into our partial equation. y=8bxx=1, y=1616=8b1 Solve for b a1=a16=8bLHS/8=RHS/82=bRearrange equation b=2 Now we can write the equation of the function represented by the table. y=8(2)x​
Exercises 9 We want to tell whether the table of values represents a linear, exponential, or quadratic function. To do so, we will analyze how the consecutive terms are related to each other.Note that the difference between consecutive x-values is always 1 — it is constant. Additionally, the consecutive y-values have a second difference of -4. This means that the table represents a quadratic function. Let's recall the intercept form of this type of function. y=a(x−p)(x−q)​ We can see in the table that there is only one x-intercept, p=q=0. Using this intercept, we can write a partial equation of the function represented by the table. y=a(x−0)(x−0)⇔y=ax2​ To find the value of a, we will substitute the ordered pair (1,-2) in the above equation. y=ax2x=1, y=-2-2=a(1)2 Solve for a 1a=1-2=a(1)Identity Property of Multiplication-2=aRearrange equation a=-2 Now we can write the equation of the function represented by the table. y=-2x2​
Exercises 10 Let's begin by considering points (-8,0) and (-2,0). Since the y-coordinate of these points is 0, we know that -8 and -2 are the x-intercepts. This means that we want to write the equation of the parabola whose x-intercepts are -8 and -2, and passes through the point (-6,4). To do so, we will use the intercept form of a quadratic function. y=a(x−p)(x−q)​ In this form, p and q are the x-intercepts. Therefore, we can already partially write our equation. y=a(x−(-8))(x−(-2))⇕y=a(x+8)(x+2)​ Finally, since the parabola passes through the point (-6,4), we can substitute -6 for x and 4 for y in our partial equation, and solve for a. y=a(x+8)(x+2)x=-6, y=44=a(-6+8)(-6+2) Solve for a Add terms4=a(2)(-4)a(-b)=-a⋅b4=a(-8)LHS/(-8)=RHS/(-8)-84​=aPut minus sign in front of fraction-84​=aba​=b/4a/4​-21​=aRearrange equation a=-21​ Knowing that a=-21​, we can write the full equation of the parabola. y=-21​(x+8)(x+2)​ Notice that this equation is written in intercept form. To write it in standard form we will use the Distributive Property. y=-21​(x+8)(x+2)Distribute (x+2)y=-21​(x(x+2)+8(x+2))Distribute xy=-21​(x2+2x+8(x+2))Distribute 8y=-21​(x2+2x+8x+16)Add termsy=-21​(x2+10x+16) Simplify right-hand side Distribute -21​y=-21​x2+(-21​)10x+(-21​)16(-a)b=-aby=-21​x2−(21​)10x−(21​)16b1​⋅a=ba​y=-21​x2−210​x−216​Calculate quotient y=-21​x2−5x−8
Exercises 11 Let's begin by considering points (0,0) and (10,0). Since the y-coordinate of these points is 0, we know that 0 and 10 are the x-intercepts. This means that we want to write the equation of the parabola whose x-intercepts are 0 and 10, and passes through the point (9,-27). To do so, we will use the intercept form of a quadratic function. y=a(x−p)(x−q)​ In this form, p and q are the intercepts. Therefore, we can already partially write our equation. y=a(x−0)(x−10)⇕y=ax(x−10)​ Finally, since the parabola passes through the point (9,-27), we can substitute 9 for x and -27 for y in our partial equation to solve for a. y=ax(x−10)x=9, y=-27-27=a(9)(9−10) Solve for a Subtract term-27=a(9)(-1)a(-b)=-a⋅b-27=a(-9)LHS/(-9)=RHS/(-9)3=aRearrange equation a=3 Knowing that a=3, we can write the full equation of the parabola. y=3x(x−10)​ Notice that this equation is written in intercept form. To write it in standard form we will use the Distributive Property. y=3x(x−10)Distribute 3xy=3x(x)−3x(10)Multiplyy=3x2−30x
Exercises 12 We are asked to find a quadratic function in standard form that is even and its range is y≥3. Let's write the standard form for a quadratic function. y=ax2+bx+c​ A function is even if and only if it is symmetric about the y-axis. Therefore the axis of symmetry of the quadratic function is equal to x=0, which is the y-axis. So, the x-coordinate of its vertex is 0. If its range is y≥3, the parabola open upwards and the y-coordinate of its vertex is equal to 3.The axis of symmetry and the range implies that the vertex is (0,3). Parabolas open upwards if a>0. Let's choose a=1 for simplicity and use the vertex form of a quadratic function. y=a(x−h)2+k​ In the formula, the point (h,k) is the vertex of the parabola. Let's substitute the values for a, h, and k, and simplify the equation. y=a(x−h)2+kSubstitute valuesy=1(x−0)2+3Subtract termy=1x2+3Multiplyy=x2+3 The equation for our quadratic function is y=x2+3. Note that this is just one solution, and there are infinitely many possible correct answers besides this.
Exercises 13 Let's start by considering the points (4,0) and (1,9). Since the y-coordinate of the first point is 0, we know that 4 is the x-intercept. It means that we want to write the equation of the parabola that passes through the point (1,9) and has x-intercept 4. To do so, we will use the intercept form of a quadratic function. y=a(x−p)(x−q)​ In this form, p and q are the intercepts. Therefore, we can partially write our equation. y=a(x−4)(x−q)​ To obtain the quadratic equation we need at least three points. Since our second given point (1,9) is not a x-intercept, we can choose any point at the x-axis to be our second x-intercept. The only point that could not be chosen is (1,0), because the obtained relation would not be a function. For simplicity, we will choose (0,0). y=a(x−4)(x−0)⇔y=ax(x−4)​ Finally, since the parabola passes through the point (1,9), we can substitute 1 for x and 9 for y in our partial equation, and solve for a. y=ax(x−4)x=1, y=99=a(1)(1−4) Solve for a Subtract term9=a(1)(-3)Identity Property of Multiplication9=a(-3)LHS/(-3)=RHS/(-3)-39​=aPut minus sign in front of fraction-39​=aCalculate quotient-3=aRearrange equation a=-3 Knowing that a=-3, we can write the full equation of the parabola. y=-3x(x−4)​ Notice that this equation is written in intercept form. To write it in standard form we need to distribute -3x. y=-3x(x−4)Distribute -3xy=-3x2+12x Please note that there are infinitely many functions that satisfy the given conditions. This is just one example.
Exercises 14 We will use patterns between consecutive data pairs to determine what type of function best models the data. The differences of consecutive y-values are called first differences. The differences of consecutive first differences are called second differences.Linear Function: The first differences are constant. Quadratic Function: The second differences are constant. Exponential Function: Consecutive y-values have a common ratio.Remember that in all cases the differences of consecutive x-values need to be constant! Let's analyze the given table and compute the first differences.Since the first differences are constant, the data can be modeled by a linear function. Next we will find a linear function that models our data. d=mt+b​ From the given table we know that the points (t,d)=(1,19) and (t,d)=(2,38) belong to the linear function d=mt+b. Let's substitute these values. {19=m(1)+b38=m(2)+b​(I)(II)​ Next, let's solve this system of equations. We will use the Substitution Method to solve this system of equations. The Substitution Method is usually the best choice when one of the variables is already isolated or has a coefficient of 1 or -1. {19=m(1)+b38=m(2)+b​(I)(II)​ Solve by substitution (I), (II): Multiply{19=m+b38=2m+b​(I)(II)​(I): LHS−m=RHS−m{19−m=b38=2m+b​(I)(II)​(II): b=19−b{19−m=b38=2m+19−m​(I)(II)​(II): Subtract term{19−m=b38=m+19​(I)(II)​(II): LHS−19=RHS−19{19−m=b19=m​(I)(II)​(I): m=19{19−19=b19=m​(I)(II)​(I): Subtract term{0=b19=m​(I)(II)​(I), (II): Rearrange equation {b=0m=19​(I)(II)​ Our data can be modeled by the linear function d=19t.
Exercises 15
Exercises 16
Exercises 17 Let's recall the formula for average rate of change of function f(x) from x=x1​ to x=x2​. Average Rage of Change=x2​−x1​f(x2​)−f(x1​)​​ Consider the given function. f(x)=x2+4​ We are asked to find average rates of change of f for three intervals.Interval f(x1​) f(x2​) x2​−x1​f(x2​)−f(x1​)​ From x1​=0 to x2​=1 02+4=4 12+4=5 1−05−4​=1 From x1​=1 to x2​=2 12+4=5 22+4=8 2−18−5​=3 From x1​=2 to x2​=3 22+4=8 32+4=13 3−213−8​=5 From the table we get that the average rate of change increases when the function f is increasing.