# How do you find the vertex and intercepts for 9x^2 – 12x + 4 = 0 ?

Apr 25, 2018

$\left(\frac{2}{3} , 0\right) \text{ and "x=2/3" (repeated)}$

#### Explanation:

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

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

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

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

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

$\text{factor out 9}$

$9 \left({x}^{2} - \frac{4}{3} x + \frac{4}{9}\right) = 0$

• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} - \frac{4}{3} x$

$9 \left({x}^{2} + 2 \left(- \frac{2}{3}\right) x \textcolor{red}{+ \frac{9}{4}} \textcolor{red}{- \frac{9}{4}} + \frac{9}{4}\right) = 0$

$\Rightarrow 9 {\left(x - \frac{2}{3}\right)}^{2} + 0 = 0$

$\text{the left side is now in "color(blue)"vertex form}$

$\text{with "h=2/3} \mathmr{and} k = 0$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{2}{3} , 0\right)$

$\text{for intercepts solve the equation }$

$\Rightarrow 9 {\left(x - \frac{2}{3}\right)}^{2} = 0$

$\Rightarrow x = \frac{2}{3} \text{( repeated)}$

$\text{this indicates a minimum at } \left(\frac{2}{3} , 0\right)$
graph{9x^2-12x+4 [-10, 10, -5, 5]}