Chapter 3.3, Problem 84AYU

Answer Problems 83 and 84 using the following: A quadratic function of the form $f\left(x\right)=a{x}^2+bx+c$ with $b^2-4ac>0$ may also be written in the form $f\left(x\right)=a(x–r_1)(x-r_2)$, where $r_1\text{ and }{r}_2$ are the $x\text{-intercepts}$ of the graph of the quadratic function.

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

(b) How does the value of $a$ 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 $x\text{-coordinate}$ of the vertex with the midpoint of the $x\text{-intercepts}$. What might you conclude?

To determine

a. The quadratic function whose $x\text{-intercepts}$ 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 $x\text{-coordinate}$ of the vertex with the midpoint of the $x\text{-intercepts}$. What might we conclude?

Answer to Problem 84AYU

a.

When $a = 1$, we have

$x^2+2x-15$

When $a = 2$, we have

$2x^2+4x-30$

When $a = -2$, we have

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

When $a = 5$, we have

$5x^2+10x-75$

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

Explanation of Solution

Given:

The $x\text{-intercepts}$ of the function are $-5$ and 3.

Formula used:

A quadratic equation of the form $f(x)=a{x}^2+bx+c,a\ne 0$ can also be written as $f(x)=a(x-r_1)(x-r_2)$ where $r_1$ and $r_2$ is the $x\text{-intercepts}$ of the graph of the quadratic function.

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

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

The $x\text{-intercept}$ 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

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

$=x^2+2x-15$

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

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

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

Vertex is at $\left(-1,-16\right)$.

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

Thus, we have

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

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

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

$and$

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

Case 2:

When $a=2$, we have

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

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

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

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

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

Vertex is at $\left(-1,-32\right)$.

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

Thus, we have

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

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

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

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

$and$

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

Case 3:

When $a=-2$, we have

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

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

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

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

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

Vertex is at $\left(-1,32\right)$.

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

Thus, we have

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

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

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

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

$and$

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

Case 4:

When $a=5$, we have

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

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

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

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

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

Vertex is at $\left(-1,-80\right)$.

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

Thus, we have

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

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

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

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

$and$

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

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 $a=1$).

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