How do you find the vertex and the intercepts for  y=3x^2+5x+9?

Jan 6, 2018

$\text{see explanation}$

Explanation:

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

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

$\text{to obtain y in this form use "color(blue)"completing the square}$

• " the coefficient of the "x^2" term must be 1"

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

• " add/subtract "(1/2"coefficient of x-term")^2" to"
${x}^{2} + \frac{5}{3} x$

$y = 3 \left({x}^{2} + 2 \left(\frac{5}{6}\right) x \textcolor{red}{+ \frac{25}{36}} \textcolor{red}{- \frac{25}{36}} + 3\right)$

$\textcolor{w h i t e}{y} = 3 {\left(x + \frac{5}{6}\right)}^{2} - \frac{25}{12} + 9$

$\textcolor{w h i t e}{y} = 3 {\left(x + \frac{5}{6}\right)}^{2} + \frac{83}{12} \leftarrow \textcolor{red}{\text{in vertex form}}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(- \frac{5}{6} , \frac{83}{12}\right)$

$\textcolor{b l u e}{\text{Intercepts}}$

• " let x = 0, in the equation for y-intercept"

• " let y = 0, in the equation for x-intercepts"

$x = 0 \to y = 3 {\left(\frac{5}{6}\right)}^{2} + \frac{83}{12} = 9 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \to 3 {\left(x + \frac{5}{6}\right)}^{2} + \frac{83}{12} = 0$

$\Rightarrow 3 {\left(x + \frac{5}{6}\right)}^{2} = - \frac{83}{12}$

$\Rightarrow {\left(x + \frac{5}{6}\right)}^{2} = - \frac{83}{36}$

$\text{this has no real solutions hence no x-intercepts}$
graph{3x^2+5x+9 [-20, 20, -10, 10]}