How do you find the vertex and intercepts for y = 3x^2 - 6x - 4?

Feb 21, 2016

vertex: $\left(1 , - 7\right)$
y-intercept: $\left(- 4\right)$
x-intercepts: $1 \pm \sqrt{\frac{7}{3}}$

Explanation:

One way to find the vertex is to convert the given equation into vertex form:
$\textcolor{w h i t e}{\text{XXX}} y = 3 {x}^{2} - 6 x - 4$

$\textcolor{w h i t e}{\text{XXX}} y = 3 \left({x}^{2} - 2 x\right) - 4$

$\textcolor{w h i t e}{\text{XXX}} y = 3 \left({x}^{2} - 2 x \textcolor{g r e e n}{+ 1}\right) \textcolor{g r e e n}{- 3} - 4$

$\textcolor{w h i t e}{\text{XXX}} y = 3 {\left(x - 1\right)}^{2} + \left(- 7\right)$
which is the vertex form for a parabola with vertex at $\left(1 , - 7\right)$

the y-intercept can be found by setting $x = 0$ in the original equation:
$\textcolor{w h i t e}{\text{XXX}} y = 3 {\left(0\right)}^{2} - 6 \left(0\right) - 4 = - 4$

The x-intercepts are a bit more work, but can be found by setting $y = 0$ in any one of the equations and solving for $x$.
I find this easiest with the vertex form above:
$\textcolor{w h i t e}{\text{XXX}} 0 = 3 {\left(x - 1\right)}^{2} - 7$

$\textcolor{w h i t e}{\text{XXX}} 3 {\left(x - 1\right)}^{2} = 7$

$\textcolor{w h i t e}{\text{XXX}} {\left(x - 1\right)}^{2} = \frac{7}{3}$

$\textcolor{w h i t e}{\text{XXX}} x - 1 = \pm \sqrt{\frac{7}{3}}$

color(white)("XXX")x=1+-sqrt(7/3)~~2.53" or "-0.53

A graph of the original equation might help show that these results are reasonable:
graph{3x^2-6x-4 [-4.13, 11.67, -7.17, 0.73]}