# How do you find the vertex and intercepts for y=4x-x^2?

Jun 15, 2018

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

#### Explanation:

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

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

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

$y = 0 \therefore$ y intercept is $y = 0 \mathmr{and} \left(0 , 0\right)$ x intercept is found by

putting $y = 0$ in the equation $y = - {x}^{2} + 4 x$ or

$0 = - {x}^{2} + 4 x \mathmr{and} x \left(4 - x\right) = 0 \therefore x = 0 \mathmr{and} x = 4$

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

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