# How do you find the vertex and intercepts for y=8(x-10)^2-16?

Sep 5, 2017

The vertex form of a quadratic function is given by
f (x) = a(x - h)2 + k, where (h, k) is the vertex of the parabola.

So here, it's already in vertex form. a = 8, h = 10, k = -16, and the vertex coordinates are 10, -16

Intercepts can be easily calculated. The y intercept is found when x = 0 is substituted into the original equation:

$y = 8 {\left(- 10\right)}^{2} - 16 = 800 - 16 = 784$

x intercepts are found by setting y = 0 and solving for x.

$0 = 8 {\left(x - 10\right)}^{2} - 16$

...easiest way to solve this is to convert it to a standard format quadratic equation.

Multiply out:

$8 \left({x}^{2} - 20 x + 100\right) - 16 = 0$

$8 {x}^{2} - 160 x + 784 = 0$

...this is standard form, $a {x}^{2} + b x + c = 0$, with
a = 8, b = -160, c = 784

giving roots 11.41 and 8.59 (rounded)

...always good to have a graph as function as a sanity check:

graph{8(x - 10)^2 - 16 [-10.51, 21.52, -19.47, -3.45]}

GOOD LUCK!

Sep 5, 2017

Vertex is $\left(10 , - 16\right)$, $x$-intercept are at $\left(10 - \sqrt{2} , 0\right)$ and $\left(10 + \sqrt{2} , 0\right)$ and $y$-intercept is at $\left(0 , 784\right)$.

#### Explanation:

To find the vertex, we should have equation into vertex form i.e. $y = a {\left(x - h\right)}^{2} + k$, in which case vertex is $\left(h , k\right)$.

As $y = 8 {\left(x - 10\right)}^{2} - 16$ is alreaddy in this form, with $a = 8$, $h = 10$ and $k = - 16$

Hence, vertex is $\left(10 , - 16\right)$

For $x$-intercepts, we should put $y = 0$ and for $y$-intercepts, we should put $x = 0$

Putting $x = 0$, $y = 8 {\left(0 - 10\right)}^{2} - 16 = 784$ and $y$-intercept is at $\left(0 , 784\right)$.

and when $y = 0$, we have $8 {\left(x - 10\right)}^{2} - 16 = 0$ i.e. ${\left(x - 10\right)}^{2} = 2$ i.e. $x = 10 \pm \sqrt{2}$ and $x$-intercept are at $\left(10 - \sqrt{2} , 0\right)$ and $\left(10 + \sqrt{2} , 0\right)$.

$\textcolor{red}{\text{Graph not drawn to scale}}$
graph{8(x-10)^2-16 [-5, 20, -100, 860]}