# How do you find the vertex and the intercepts for y = -4x^2 + 16x - 3?

Apr 12, 2018

Vertex is $\left(2 , 13\right)$

#### Explanation:

The vertex of the parabola will be the at he maximum of the

function,i.e, where $\frac{\mathrm{dy}}{\mathrm{dx}} = 0.$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 8 x + 16 = 0$

$\implies x = 2$
The corresponding value of $y$ is $13$

The $y$-intercept will be where $x = 0$ and the $x$-intercept will be where $y = 0$

So, $y$-intercept $= - 3$
x-intercept is the solution of the equation

$\text{ } - 4 {x}^{2} + 16 x - 3 = 0$

which by solving we get $x = 0.197$ and $x = 3.803$ , these are the $x$-intercepts.
graph{-4x^2+16x-3 [-10.82, 21.23, -0.42, 15.6]}
All these can be checked from the graph of the function

.