# How do you find the vertex and the intercepts for G(x) = x^2 + 4x?

Dec 10, 2017

Vertex is at $\left(- 2 , - 4\right)$ , y intercept is at $\left(0 , 0\right)$ and
x intercepts are at
$\left(0 , 0\right) \mathmr{and} \left(- 4 , 0\right)$

#### Explanation:

$G \left(x\right) = {x}^{2} + 4 x \mathmr{and} G \left(x\right) = {x}^{2} + 4 x + 4 - 4$ or

$G \left(x\right) = {\left(x + 2\right)}^{2} - 4$ . Comparing with vertex form of

equation f(x) = a(x-h)^2+k ; (h,k) being vertex we find

here $h = - 2 , k = - 4 \therefore$ Vertex is at $\left(- 2 , - 4\right)$ .

y intercept is found by putting $x = 0$ in the equation

$G \left(x\right) = {x}^{2} + 4 x$ or $y = 0 \therefore$ y intercept is $\left(0 , 0\right)$

x intercepts are found by putting $y = 0$ in the equation

$G \left(x\right) = y = {x}^{2} + 4 x \mathmr{and} {x}^{2} + 4 x = 0$ or

$x \left(x + 4\right) = 0 \therefore x = 0 \mathmr{and} x + 4 = 0 \therefore x = - 4$

x intercepts are at $\left(0 , 0\right) \mathmr{and} \left(- 4 , 0\right)$

graph{x^2+4x [-10, 10, -5, 5]} [Ans]