Question

The zeros of a quadratic function are (-2,0) and (8,0). What is a possible vertex of the function? A. (3,0)B. (3,-1)C. (6,0)D. (6,-1)

Answer 1

B(3,-1)

Explanation

$y=ax^2+bx+c$

Step 1

Point 1

(-2,0)

replace

$\begin{gathered} y=ax^2+bx+c \\ 0=a(-2)^2+b(-2)+c \\ 0=4a-2b+c \\ 2b-4a=c\rightarrow equation\text{ (1)} \end{gathered}$

Point 2

(8,0)

$\begin{gathered} y=ax^2+bx+c \\ 0=a(8)^2+b(8)+c \\ 0=64a+8b+c\rightarrow equation\text{ (2)} \\ c=-64a-8b \end{gathered}$

c) c= c, so

$\begin{gathered} 2b-4a=-64a-8b \\ -4a+64a=-8b-2b \\ 60a=-10b \\ \frac{60a}{-10}=\frac{-10b}{-10} \\ b=-6a \end{gathered}$

Step 2

when the function is in the form

$y=ax^2+bx+c$

the vertex is given by

$(-\frac{b}{2a},f(-\frac{b}{2a})$

so,replace

$\begin{gathered} -\frac{b}{2a} \\ -\frac{(-6a)}{2a}=\frac{6a}{2a}=3 \\ \end{gathered}$

therefore, the component of the vertex is 3

let's check the options

A. (3,0)

B. (3,-1)

we can see that y component of option A is zero , it means (3,0 ) is a zero of the function, but we had already the zeros, therefore, we can discard this options,

in other words,

the answer is

B(3,-1)

I hope this helps you