Question

The y coordinate of the vertex of f (x)=-x^2+8x+7 is f(4)

Answer 1

Hello there.

To answer this question, we need to remember some properties about the vertex of quadratic functions.

Let $f(x)=ax^2+bx+c,~a\neq0$. Its vertex can be found on the coordinates $(x_v,~y_v)$ such that $x_v=-\dfrac{b}{2a}$ and $y_v=-\dfrac{b^2}{4a}+c$, in which $y_v=f(x_v)$.

Using the coefficients given by the question, we get that:

$x_v=-\dfrac{8}{2\cdot(-1)}\\\\\\ x_v=-\dfrac{8}{-2}\\\\\\ x_v=4$

Thus, we have:

$y=f(x_v)=f(4)$

So the statement is true, because the $x$ coordinate of the vertex of the function is equal to $4.~~\checkmark$