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:

Step-by-step explanation:

The x-intercepts are the points with y- coordinate of 0

a) x-intercepts are

b) midpoint of x-intercepts is

c) The extheme value has x- coordinate of - 2.5 and y- coordinate

d) f(- 2.5) = - 30.25

Answer:

[tex]\sf a) \quad x = -8 \:\:and \:\:x = 3[/tex]

[tex]\sf b) \quad \left(-\dfrac{5}{2},0\right)[/tex]

[tex]\sf c) \quad \left(-\dfrac{5}{2},-\dfrac{121}{4}\right)[/tex]

[tex]\sf d) \quad f(x)=\left(x+\dfrac{5}{2}\right)^2-\dfrac{121}{4}[/tex]

Step-by-step explanation:

Part (a)

The x-intercepts are the x-values when the curve crosses the x-axis, so when the function is equal to zero:

[tex]\implies f(x)=0[/tex]

[tex]\implies (x-3)(x+8)=0[/tex]

Therefore:

[tex](x-3)=0 \implies x=3[/tex]

[tex](x+8)=0 \implies x=-8[/tex]

Therefore, the x-intercepts are x = -8 and x = 3

Part (b)

[tex]\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)[/tex]

[tex]\sf \implies Midpoint=\left(\dfrac{-8+3}{2},\dfrac{0+0}{2}\right)=\left(-\dfrac{5}{2},0\right)[/tex]

Therefore, the midpoint of the x-intercepts is (-5/2, 0)

Part (c)

The extreme value is the vertex.  The vertex is:

The x-coordinate of the vertex is the x-value of the midpoint of the x-intercepts.  Therefore, the x-value of the extreme value is x = -5/2.

To find the y-coordinate, substitute this into the given equation:

[tex]\implies \sf y=\left(-\dfrac{5}{2}-3\right)\left(-\dfrac{5}{2}+8\right)=-\dfrac{121}{4}[/tex]

Therefore, the extreme value is:

[tex]\sf \left(-\dfrac{5}{2},-\dfrac{121}{4}\right)[/tex]

Part (d)

The function in vertex form is:

[tex]\sf f(x)=\left(x+\dfrac{5}{2}\right)^2-\dfrac{121}{4}[/tex]

[tex]\begin{aligned}\implies \sf f\left(-\dfrac{5}{2}\right)& = \sf \left(-\dfrac{5}{2}+\dfrac{5}{2}\right)^2-\dfrac{121}{4}\\\\ & = \sf 0 -\dfrac{121}{4}\\\\ & = \sf -\dfrac{121}{4} \end{aligned}[/tex]

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