# How do you find the vertex and intercepts for y=3x^2+5x+8?

Jul 4, 2018

$\text{see explanation}$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

•color(white)(x)y=a(x-h)^2+k

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{to obtain this form "color(blue)"complete the square}$

$y = 3 \left({x}^{2} + \frac{5}{3} x + \frac{8}{3}\right)$

$\textcolor{w h i t e}{y} = 3 \left({x}^{2} + 2 \left(\frac{5}{6}\right) x \textcolor{red}{+ \frac{25}{36}} \textcolor{red}{- \frac{25}{36}} + \frac{8}{3}\right)$

$\textcolor{w h i t e}{y} = 3 {\left(x + \frac{5}{6}\right)}^{2} + 3 \left(- \frac{25}{36} + \frac{96}{36}\right)$

color(white)(y)=3(x+5/6)^2+71/12larrcolor(blue)"in vertex form"

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- \frac{5}{6} , \frac{71}{12}\right)$

$\text{to find the x-intercepts let y = 0}$

$3 {\left(x + \frac{5}{6}\right)}^{2} + \frac{71}{12} = 0$

$3 {\left(x + \frac{5}{6}\right)}^{2} = - \frac{71}{12}$

$\text{this has no real solutions hence no x-intercepts}$

$\text{for y-intercept let x = 0}$

$y = 0 + 0 + 8 = 8 \leftarrow \textcolor{red}{\text{y-intercept}}$
graph{3x^2+5x+8 [-20, 20, -10, 10]}