# How do you find the vertex, x and y intercepts for f(x) = -3x^2 + 6x -2?

Jan 18, 2016

Vertex ->(x.y)->(+1,?)
Having shown you a cool trick to get the value of $x$ I will let you finish it to find $y$

#### Explanation:

color(blue)("To find "x_("vertex"))

Given: $\textcolor{w h i t e}{\text{xx}} y = - 3 {x}^{2} + 6 x - 2 \textcolor{w h i t e}{.}$ .......(1)

Write as: $- 3 \left({x}^{2} \textcolor{b r o w n}{- 2} x\right) - 2$

Now consider the $\textcolor{b r o w n}{- 2} \text{ from the "-2x" inside the brackets}$

Multiply this value by $\left(- \frac{1}{2}\right)$

$\left(- \frac{1}{2}\right) \times \left(- 2\right) = + 1$

$\textcolor{b l u e}{{x}_{\text{vertex}} = + 1}$
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Compare this to the graph:

Ok! You have now found the value for $x = 1$

Substitute this back into equation (1) to find the value for $y$
I will let you do that!

Jan 18, 2016

Vertex (1, 1)
y = -2
$x = \frac{3 \pm \sqrt{3}}{3}$

#### Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = - \frac{6}{-} 6 = 1$
y-coordinate of vertex:
$f \left(1\right) = - 3 + 6 - 2 = 1.$
Vertex (1, 1)
To find y-intercept, make x = 0 --> y = -2
To find x-intercepts, make y = 0 and solve:
$f \left(x\right) = - 3 {x}^{2} + 6 x - 2 = 0$
$D = {d}^{2} = {b}^{2} - 4 a c = 36 - 24 = 12$ --> $d = \pm 2 \sqrt{3}$
$x = - \frac{b}{2 a} \pm \frac{d}{2 a} = 1 \pm \frac{\sqrt{3}}{3} = \frac{3 \pm \sqrt{3}}{3}$