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

(a) (3, 0) and (-8, 0)

(b) (-2.5, 0)

(c) x = -2.5

(d) f(-2.5) = -30.25

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((x_2+x_1)/(2),(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((-8+3)/(2),(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=-(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=-(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}`