Question

Use the method of symmetry to find the extreme value of each quadratic function and the value of x for which it occurs.
f(x)=(x-3)(x+8

a) x-intercepts are _________
b) midpoint of the x-intercepts is ___________
c) the extreme value is _______________
d) f(_) = ______________

Answer 1

Answer:

(a)  (3, 0) and (-8, 0)

(b)  (-2.5, 0)

(c) x = -2.5

(d) f(-2.5) = -30.25

Step-by-step explanation:

Given quadratic function:

$f(x)=(x-3)(x+8)$

Part (a)

The x-intercepts are when $f(x)=0$

$\implies f(x)=0$

$\implies (x-3)(x+8)=0$

$\implies (x-3)=0 \implies x=3$

$\implies (x+8)=0 \implies x=-8$

Therefore, the x-intercepts are (3, 0) and (-8, 0)

Part (b)

Midpoint between two points:

$\textsf{Midpoint}=\left(\dfrac{x_2+x_1}{2},\dfrac{y_2+y_1}{2}\right)\quad$

$\textsf{where}\:(x_1,y_1)\:\textsf{and}\:(x_2,y_2)\:\textsf{are the endpoints}}\right)$

$\implies \textsf{Midpoint of the x-intercepts}=\left(\dfrac{-8+3}{2},\dfrac{0+0}{2}\right)=\left(-2.5},0\right)$

Part (c)

The extreme point of a quadratic function in the form $f(x)=ax^2+bx+c$ is:

$x=-\dfrac{b}{2a}$

Therefore, expand the function so that it is in standard form:

$\implies f(x)=x^2+5x-24$

$\implies a=1, b=5$

Therefore, the extreme value is:

$\implies x=-\dfrac{5}{2}=-2.5$

Alternative method

A quadratic function has an extreme value at its vertex.

The x-value of the vertex is the midpoint of the x-intercepts.

Therefore, the extreme value is x = -2.5

Part (d)

$\begin{aligned}\implies f(-2.5) & =(-2.5-3)(-2.5+8)\\& = (-5.5)(5.5)\\& = -30.25\end{aligned}$