Question

Hello can you please help me out on this question posted picture

Answer 1

The answer is A. I calculated it and the vertex came out 2, -25 and the intercept was -3,7

Answer 2

First we need to find the vertex.
Vertex of a quadratic functions exists at:
$( \frac{-b}{2a},y( \frac{-b}{2a}))$
b = coefficient of x term = -4a = coefficient of squared term = 1
So,
$\frac{-b}{2a}= \frac{4}{2}=2$
and
$y(2)=(2)^{2}-4(2)-21=-25$
Therefore, the vertex occurs at (2, -25)

x-intercepts are the points where the value of function is zero.

So,

$x^{2} -4x-21=0 \\ \\ x^{2} -7x+3x-21=0 \\ \\ x(x-7)+3(x-7)=0 \\ \\ (x-7)(x+3)=0 \\ \\ x=7, x=-3$

Thus, the x-intercepts are (7,0) and (-3,0)

So option A gives the correct answer