Answer
a) $x = -1$
b) $x = 2, -4$
c) $x = 0, -2$
So $(0, -8)$ and $(-2, -8)$ lies on the graph
d) see graph
Work Step by Step
$f(x) = x^2 + 2x - 8$
$a = 1$, $b = 2$ $c = -8$
a)
Vertex of f(x)
$x = \frac{-b}{2a} = -1$
b)
x-intercepts - Solution $f(x) = 0$ will give x-intercepts
$\Rightarrow x^2 + 2x - 8 = 0$
$\Rightarrow (x+4)(x-2) = 0$
$\Rightarrow x = 2, -4$
c)
$f(x) = -8$
$\Rightarrow x^2 + 2x - 8 = -8$
$\Rightarrow x(x+2) = 0$
$\Rightarrow x = 0, -2$
So $(0, -8)$ and $(-2, -8)$ lies on the graph
