# What are the important points needed to graph y = 8(x - 3)^2 - 5?

Jul 3, 2018

vertex $\left(3 , - 5\right)$
$x$-intercepts: $\left(3 - \frac{\sqrt{10}}{4} , 0\right) , \left(3 + \frac{\sqrt{10}}{4} , 0\right)$

$y$-intercept: (0, 67)

#### Explanation:

Given: $y = 8 {\left(x - 3\right)}^{2} - 5$

$\textcolor{b l u e}{\text{Find the vertex:}}$

From the vertex form: $y = a {\left(x - h\right)}^{2} + k$, where vertex $\left(h , k\right)$

we can find the vertex of the given equation as $\left(3 , - 5\right)$

$\textcolor{b l u e}{\text{Find the "y-"intercept}}$ by setting $x = 0$:

$y = 8 {\left(0 - 3\right)}^{2} - 5$

$y = 8 \cdot 9 - 5 = 72 - 5 = 67$

$y$-intercept: (0, 67)

$\textcolor{b l u e}{\text{Find the "x-"intercepts}}$ by setting $y = 0$:

$0 = 8 {\left(x - 3\right)}^{2} - 5$

Distribute using the square function: ${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

$8 \left({x}^{2} - 6 x + 9\right) - 5 = 0$

$8 {x}^{2} - 48 x + 72 - 5 = 0$

$8 {x}^{2} - 48 x + 67 = 0$

Use the quadratic formula to solve:

$x = \frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A}$,

where the equation is in the form: $A {x}^{2} + B x + C = 0$

$x = \frac{48 \pm \sqrt{{\left(- 48\right)}^{2} - 4 \cdot 8 \cdot 67}}{16}$

$x = \frac{48}{16} \pm \frac{\sqrt{160}}{16}$

$x = 3 \pm \frac{\sqrt{16 \cdot 10}}{16}$

$x = 3 \pm \frac{4 \sqrt{10}}{16}$

$x = 3 \pm \frac{\sqrt{10}}{4}$

$x$-intercepts: $\left(3 - \frac{\sqrt{10}}{4} , 0\right) , \left(3 + \frac{\sqrt{10}}{4} , 0\right)$