# How do you find the vertex and the intercepts for f(x) = 3x^2 -12x +10?

Apr 18, 2016

Use a few formulas to find the vertex is $\left(2 , - 2\right)$ and the intercepts are $\left(2 + \frac{\sqrt{6}}{3} , 0\right)$ and $\left(2 - \frac{\sqrt{6}}{3} , 0\right)$.

#### Explanation:

The $x$-coordinate of the vertex of the parabola $a {x}^{2} + b x + c$ is:
$x = - \frac{b}{2 a}$

In our case $a = 3$ and $b = - 12$, which means the $x$-coordinate of the vertex is $- \frac{- 12}{2 \left(3\right)} = 2$. The $y$-coordinate can be found by evaluating $3 {x}^{2} - 12 x + 10$ at $x = 2$:
$y = 3 {x}^{2} - 12 x + 10 = 3 {\left(2\right)}^{2} - 12 \left(2\right) + 10 = - 2$

Thus the vertex is $\left(2 , - 2\right)$.

Finding the intercepts will require use of the quadratic formula, which is:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

We have $a = 3$, $b = - 12$, and $c = 10$:
$x = \frac{- \left(- 12\right) \pm \sqrt{{\left(- 12\right)}^{2} - 4 \left(3\right) \left(10\right)}}{2 \left(3\right)}$
$\textcolor{w h i t e}{X X} = \frac{12 \pm \sqrt{144 - 120}}{6}$
$\textcolor{w h i t e}{X X} = 2 \pm \frac{\sqrt{24}}{6}$
$\textcolor{w h i t e}{X X} = 2 \pm \frac{2 \sqrt{6}}{6}$
$\textcolor{w h i t e}{X X} = 2 \pm \frac{\sqrt{6}}{3}$

The intercepts are $\left(2 + \frac{\sqrt{6}}{3} , 0\right)$ and $\left(2 - \frac{\sqrt{6}}{3} , 0\right)$.