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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:
[tex]f(x)=(x-3)(x+8)[/tex]
The x-intercepts are when [tex]f(x)=0[/tex]
[tex]\implies f(x)=0[/tex]
[tex]\implies (x-3)(x+8)=0[/tex]
[tex]\implies (x-3)=0 \implies x=3[/tex]
[tex]\implies (x+8)=0 \implies x=-8[/tex]
Therefore, the x-intercepts are (3, 0) and (-8, 0)
Midpoint between two points:
[tex]\textsf{Midpoint}=\left(\dfrac{x_2+x_1}{2},\dfrac{y_2+y_1}{2}\right)\quad[/tex]
[tex]\textsf{where}\:(x_1,y_1)\:\textsf{and}\:(x_2,y_2)\:\textsf{are the endpoints}}\right)[/tex]
[tex]\implies \textsf{Midpoint of the x-intercepts}=\left(\dfrac{-8+3}{2},\dfrac{0+0}{2}\right)=\left(-2.5},0\right)[/tex]
The extreme point of a quadratic function in the form [tex]f(x)=ax^2+bx+c[/tex] is:
[tex]x=-\dfrac{b}{2a}[/tex]
Therefore, expand the function so that it is in standard form:
[tex]\implies f(x)=x^2+5x-24[/tex]
[tex]\implies a=1, b=5[/tex]
Therefore, the extreme value is:
[tex]\implies x=-\dfrac{5}{2}=-2.5[/tex]
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
[tex]\begin{aligned}\implies f(-2.5) & =(-2.5-3)(-2.5+8)\\& = (-5.5)(5.5)\\& = -30.25\end{aligned}[/tex]