To determine a. The quadratic function whose x -intercepts is given. b. How does the value a affect the intercepts? c. How does the value of a affect the axis of symmetry? d. How does the value of a affect the vertex? e. Compare the x -coordinate of the vertex with the midpoint of the x -intercepts . What might we conclude? Answer a. When a = 1 , we have x 2 + 2 x − 15 When a = 2 , we have 2 x 2 + 4 x − 30 When a = − 2 , we have − 2 x 2 − 4 x + 30 When a = 5 , we have 5 x 2 + 10 x − 75 b. We can see that the value of a does not affect the intercepts. c. The value of a does not affect the axis of symmetry. d. As the value of a increases the y -value of vertex is decreasing a times (when compared with the quadratic function at a = 1 ). e. The midpoint of the given x -intercepts is at − 5 + 3 2 = − 1 . We can see that the x -coordinate of the vertex is same as the midpoint of the x -intercepts . Thus, we can conclude that the x -coordinate of the vertex and the axis of symmetry are all same as the midpoint of the x -intercepts of the function. Explanation Given: The x -intercepts of the function are − 5 and 3. Formula used: A quadratic equation of the form f ( x ) = a x 2 + b x + c , a ≠ 0 can also be written as f ( x ) = a ( x − r 1 ) ( x − r 2 ) where r 1 and r 2 is the x -intercepts of the graph of the quadratic function. Axis of symmetry is x = − b 2 a Vertex is at ( − b 2 a , f ( − b 2 a ) ) The x -intercept is found by solving the equation at f ( x ) = 0 . Calculation: a. The given quadratic function is f ( x ) = a ( x − ( − 5 ) ) ( x − 3 ) ⇒ f ( x ) = a ( x + 5 ) ( x − 3 ) Case 1: When a = 1 , we have f ( x ) = 1 ( x + 5 ) ( x − 3 ) = x 2 + 2 x − 15 Here, we get a = 1 , b = 2 , c = − 15 Axis of symmetry is x = − 2 2 ( 1 ) = − 1 Thus, we have f ( − b 2 a ) = f ( − 1 ) = ( − 1 ) 2 + 2 ( − 1 ) − 15 = − 16 Vertex is at ( − 1 , − 16 ) . The x -intercept is found by solving the equation at f ( x ) = 0 . Thus, we have x 2 + 2 x − 15 = 0 ⇒ ( x + 5 ) ( x − 3 ) = 0 ⇒ x + 5 = 0 ⇒ x = − 5 a n d ⇒ x − 3 = 0 ⇒ x = 3 Case 2: When a = 2 , we have f ( x ) = 2 ( x + 5 ) ( x − 3 ) = 2 x 2 + 4 x − 30 Here, we get a = 2 , b = 4 , c = − 30 Axis of symmetry is x = − 2 2 ( 1 ) = − 1 Thus, we have f ( − b 2 a ) = f ( − 1 ) = 2 ( − 1 ) 2 + 4 ( − 1 ) − 30 = − 32 Vertex is at ( − 1 , − 32 ) . The x -intercept is found by solving the equation at f ( x ) = 0 . Thus, we have 2 x 2 + 4 x − 30 = 0 ⇒ 2 ( x 2 + 2 x − 15 ) = 0 ⇒ 2 ( x + 5 ) ( x − 3 ) = 0 ⇒ x + 5 = 0 ⇒ x = − 5 a n d ⇒ x − 3 = 0 ⇒ x = 3 Case 3: When a = − 2 , we have f ( x ) = − 2 ( x + 5 ) ( x − 3 ) = − 2 x 2 − 4 x + 30 Here, we get a = − 2 , b = − 4 , c = 30 Axis of symmetry is x = − 2 2 ( 1 ) = − 1 Thus, we have f ( − b 2 a ) = f ( − 1 ) = − 2 ( − 1 ) 2 − 4 ( − 1 ) + 30 = 32 Vertex is at ( − 1 , 32 ) . The x -intercept is found by solving the equation at f ( x ) = 0 . Thus, we have 2 x 2 + 4 x − 30 = 0 ⇒ 2 ( x 2 + 2 x − 15 ) = 0 ⇒ 2 ( x + 5 ) ( x − 3 ) = 0 ⇒ x + 5 = 0 ⇒ x = − 5 a n d ⇒ x − 3 = 0 ⇒ x = 3 Case 4: When a = 5 , we have f ( x ) = 5 ( x + 5 ) ( x − 3 ) = 5 x 2 + 10 x − 75 Here, we get a = 5 , b = 10 , c = − 75 Axis of symmetry is x = − 2 2 ( 1 ) = − 1 Thus, we have ( − b 2 a ) = f ( − 1 ) = 5 ( − 1 ) 2 + 10 ( − 1 ) − 75 = − 80 Vertex is at ( − 1 , − 80 ) . The x -intercept is found by solving the equation at f ( x ) = 0 . Thus, we have 5 x 2 + 10 x − 75 = 0 ⇒ 5 ( x 2 + 2 x − 15 ) = 0 ⇒ 5 ( x + 5 ) ( x − 3 ) = 0 ⇒ x + 5 = 0 ⇒ x = − 5 a n d ⇒ x − 3 = 0 ⇒ x = 3 b. We can see that the value of a does not affect the intercepts. c. The value of a does not affect the axis of symmetry. d. As the value of a increases the y -value of vertex is decreasing a times (when compared with the quadratic function at a = 1 ). e. The midpoint of the given x -intercepts is at − 5 + 3 2 = − 1 . We can see that the x -coordinate of the vertex is same as the midpoint of the x -intercepts . Thus, we can conclude that the x -coordinate of the vertex and the axis of symmetry are all same as the midpoint of the x -intercepts of the function.

Precalculus Enhanced with Graphing Utilities

6th Edition

ISBN: 9780321795465

Author: Michael Sullivan, Michael III Sullivan

Publisher: PEARSON

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Textbook Question

Chapter 3.3, Problem 84AYU

Answer Problems 83 and 84 using the following: A quadratic function of the form $f (x) = a x^{2} + b x + c$ with $b^{2} - 4 a c > 0$ may also be written in the form $f (x) = a (x - r_{1}) (x - r_{2})$ , where $r_{1} and r_{2}$ are the $x -intercepts$ of the graph of the quadratic function.

(a) Find a quadratic function whose $x -intercepts$ are $- 5$ and 3 with $a = 1; a = 2; a = - 2; a = 5$ .

(b) How does the value of $a$ affect the intercepts?

(d) How does the value of $a$ affect the vertex?

(e) Compare the $x -coordinate$ of the vertex with the midpoint of the $x -intercepts$ . What might you conclude?

Expert Solution & Answer

To determine

a. The quadratic function whose $x -intercepts$ is given.

b. How does the value $a$ affect the intercepts?

c. How does the value of $a$ affect the axis of symmetry?

d. How does the value of $a$ affect the vertex?

e. Compare the $x -coordinate$ of the vertex with the midpoint of the $x -intercepts$ . What might we conclude?

Answer to Problem 84AYU

When $a = 1$ , we have

$x^{2} + 2 x - 15$

When $a = 2$ , we have

$2 x^{2} + 4 x - 30$

When $a = - 2$ , we have

$- 2 x^{2} - 4 x + 30$

When $a = 5$ , we have

$5 x^{2} + 10 x - 75$

b. We can see that the value of $a$ does not affect the intercepts.

c. The value of $a$ does not affect the axis of symmetry.

d. As the value of $a$ increases the $y$ -value of vertex is decreasing $a$ times (when compared with the quadratic function at $a = 1$ ).

e. The midpoint of the given $x -intercepts$ is at $\frac{- 5 + 3}{2} = - 1$ . We can see that the $x -coordinate$ of the vertex is same as the midpoint of the $x -intercepts$ . Thus, we can conclude that the $x -coordinate$ of the vertex and the axis of symmetry are all same as the midpoint of the $x -intercepts$ of the function.

Explanation of Solution

Given:

The $x -intercepts$ of the function are $- 5$ and 3.

Formula used:

A quadratic equation of the form $f (x) = a x^{2} + b x + c, a \neq 0$ can also be written as $f (x) = a (x - r_{1}) (x - r_{2})$ where $r_{1}$ and $r_{2}$ is the $x -intercepts$ of the graph of the quadratic function.

Axis of symmetry is $x = \frac{- b}{2 a}$

Vertex is at $(\frac{- b}{2 a}, f (\frac{- b}{2 a}))$

The $x -intercept$ is found by solving the equation at $f (x) = 0$ .

Calculation:

a. The given quadratic function is

$f (x) = a (x - (- 5)) (x - 3)$

$\Rightarrow f (x) = a (x + 5) (x - 3)$

Case 1:

When $a = 1$ , we have

$f (x) = 1 (x + 5) (x - 3)$

$= x^{2} + 2 x - 15$

Here, we get $a = 1, b = 2, c = - 15$

Axis of symmetry is $x = \frac{- 2}{2 (1)} = - 1$

Thus, we have $f (\frac{- b}{2 a}) = f (- 1) = {(- 1)}^{2} + 2 (- 1) - 15 = - 16$

Vertex is at $(- 1, - 16)$ .

The $x -intercept$ is found by solving the equation at $f (x) = 0$ .

Thus, we have

$x^{2} + 2 x - 15 = 0$

$\Rightarrow (x + 5) (x - 3) = 0$

$\Rightarrow x + 5 = 0 \Rightarrow x = - 5$

$a n d$

$\Rightarrow x - 3 = 0 \Rightarrow x = 3$

Case 2:

When $a = 2$ , we have

$f (x) = 2 (x + 5) (x - 3)$

$= 2 x^{2} + 4 x - 30$

Here, we get $a = 2, b = 4, c = - 30$

Axis of symmetry is $x = \frac{- 2}{2 (1)} = - 1$

Thus, we have $f (\frac{- b}{2 a}) = f (- 1) = 2 {(- 1)}^{2} + 4 (- 1) - 30 = - 32$

Vertex is at $(- 1, - 32)$ .

The $x -intercept$ is found by solving the equation at $f (x) = 0$ .

Thus, we have

$2 x^{2} + 4 x - 30 = 0$

$\Rightarrow 2 (x^{2} + 2 x - 15) = 0$

$\Rightarrow 2 (x + 5) (x - 3) = 0$

$\Rightarrow x + 5 = 0 \Rightarrow x = - 5$

$a n d$

$\Rightarrow x - 3 = 0 \Rightarrow x = 3$

Case 3:

When $a = - 2$ , we have

$f (x) = - 2 (x + 5) (x - 3)$

$= - 2 x^{2} - 4 x + 30$

Here, we get $a = - 2, b = - 4, c = 30$

Axis of symmetry is $x = \frac{- 2}{2 (1)} = - 1$

Thus, we have $f (\frac{- b}{2 a}) = f (- 1) = - 2 {(- 1)}^{2} - 4 (- 1) + 30 = 32$

Vertex is at $(- 1, 32)$ .

The $x -intercept$ is found by solving the equation at $f (x) = 0$ .

Thus, we have

$2 x^{2} + 4 x - 30 = 0$

$\Rightarrow 2 (x^{2} + 2 x - 15) = 0$

$\Rightarrow 2 (x + 5) (x - 3) = 0$

$\Rightarrow x + 5 = 0 \Rightarrow x = - 5$

$a n d$

$\Rightarrow x - 3 = 0 \Rightarrow x = 3$

Case 4:

When $a = 5$ , we have

$f (x) = 5 (x + 5) (x - 3)$

$= 5 x^{2} + 10 x - 75$

Here, we get $a = 5, b = 10, c = - 75$

Axis of symmetry is $x = \frac{- 2}{2 (1)} = - 1$

Thus, we have $(\frac{- b}{2 a}) = f (- 1) = 5 {(- 1)}^{2} + 10 (- 1) - 75 = - 80$

Vertex is at $(- 1, - 80)$ .

The $x -intercept$ is found by solving the equation at $f (x) = 0$ .

Thus, we have

$5 x^{2} + 10 x - 75 = 0$

$\Rightarrow 5 (x^{2} + 2 x - 15) = 0$

$\Rightarrow 5 (x + 5) (x - 3) = 0$

$\Rightarrow x + 5 = 0 \Rightarrow x = - 5$

$a n d$

$\Rightarrow x - 3 = 0 \Rightarrow x = 3$

b. We can see that the value of $a$ does not affect the intercepts.

c. The value of $a$ does not affect the axis of symmetry.

d. As the value of $a$ increases the $y$ -value of vertex is decreasing $a$ times (when compared with the quadratic function at $a = 1$ ).

Chapter 3.3, Problem 83AYU

Chapter 3.3, Problem 85AYU

Chapter 3 Solutions

Precalculus Enhanced with Graphing Utilities

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Solution Summary: The author explains that the x-coordinate of the vertex and axis of symmetry are all the same.

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Answer to Problem 84AYU

Explanation of Solution

Chapter 3 Solutions

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