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

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Answer 1 · 2017-12-10T11:39:05

Vertex is at $(-2,-4)$ , y intercept is at $(0,0)$ and
x intercepts are at $(0,0) and (-4,0)$

$G(x)=x^2+4x or G(x)=x^2+4x+4-4$ or

$G(x)=(x+2)^2-4$ . Comparing with vertex form of

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

here $h=-2 , k=-4 :.$ Vertex is at $(-2,-4)$ .

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

$G(x)=x^2+4x$ or $y=0 :.$ y intercept is $(0,0)$

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

$G(x)=y=x^2+4x or x^2+4x =0$ or

$x(x+4) =0 :. x=0 and x+4=0 :. x =-4$

x intercepts are at $(0,0) and (-4,0)$

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

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