# What is the axis of symmetry and vertex for the graph y = 3x^2 - 9x + 12?

Jul 5, 2018

$x = \frac{3}{2} , \text{ vertex } = \left(\frac{3}{2} , \frac{21}{4}\right)$

#### Explanation:

$\text{given a quadratic in "color(blue)"standard form}$

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

$\text{then the axis of symmetry which is also the x-coordinate}$
$\text{of the vertex is}$

$\textcolor{w h i t e}{x} {x}_{\textcolor{red}{\text{vertex}}} = - \frac{b}{2 a}$

$y = 3 {x}^{2} - 9 x + 12 \text{ is in standard form}$

$\text{with "a=3,b=-9" and } c = 12$

${x}_{\text{vertex}} = - \frac{- 9}{6} = \frac{3}{2}$

$\text{substitute this value into the equation for y-coordinate}$

${y}_{\text{vertex}} = 3 {\left(\frac{3}{2}\right)}^{2} - 9 \left(\frac{3}{2}\right) + 12 = \frac{21}{4}$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{3}{2} , \frac{21}{4}\right)$

$\text{equation of axis of symmetry is } x = \frac{3}{2}$
graph{(y-3x^2+9x-12)((x-3/2)^2+(y-21/4)^2-0.04)=0 [-14.24, 14.24, -7.12, 7.12]}

$x = \frac{3}{2}$ & $\left(\frac{3}{2} , \frac{21}{4}\right)$

#### Explanation:

Given equation:

$y = 3 {x}^{2} - 9 x + 12$

$y = 3 \left({x}^{2} - 3 x\right) + 12$

$y = 3 \left({x}^{2} - 3 x + \frac{9}{4}\right) - \frac{27}{4} + 12$

$y = 3 {\left(x - \frac{3}{2}\right)}^{2} + \frac{21}{4}$

${\left(x - \frac{3}{2}\right)}^{2} = \frac{1}{3} \left(y - \frac{21}{4}\right)$

The above equation shows an upward parabola: ${X}^{2} = 4 A Y$ which has

Axis of symmetry : $X = 0 \setminus \implies x - \frac{3}{2} = 0$

$x = \frac{3}{2}$

Vertex: $\left(X = 0 , Y = 0\right) \setminus \equiv \left(x - \frac{3}{2} = 0 , y - \frac{21}{4} = 0\right)$

$\left(\frac{3}{2} , \frac{21}{4}\right)$