# How do you find the vertex and the intercepts for -3x^2+12x-9?

May 7, 2016

vertex: maximum point(2,3)
Y-intercept = -9
X-intercept = 1 and 3

#### Explanation:

The given equation is $- 3 {x}^{2} + 12 x - 9 = y$.
Given that the coefficient(constant number in front) of ${x}^{2}$ is negative 3, for any NEGATIVE coefficient, the graph is a SAD smiley (has a maximum point).
Note for a POSITIVE coefficient, the graph is a HAPPY smiley (has a minimum point).

Complete the square to find the vertex:

$- 3 {x}^{2} + 12 x - 9 = 0$
$- 3 \left({x}^{2} - 4 x\right) - 9 = 0$
$- 3 {\left(x - 2\right)}^{2} - \left(- 3\right) {\left(2\right)}^{2} - 9 = 0$
$- 3 {\left(x - 2\right)}^{2} + 12 - 9 = 0$
$- 3 {\left(x - 2\right)}^{2} + 3 = 0$

In this form $- 3 {\left(x - 2\right)}^{2} + 3 = 0$,
the constant 3 is the y coordinate of the vertex.
And (x-2) = 0 is the x coordinate of the vertex, equate for x,
x=2 is the x coordinate. Hence the vertex is (2, 3).

To find the intercepts:

For y-intercept, equate x=0,
$- 3 {\left(0\right)}^{2} + 12 \left(0\right) - 9 = y$
$0 + 0 - 9 = y$
y= -9 is the y coordinate of the y-intercept.

For x-intercept, equate y=0, then factorise to find the 2 x intercepts,
$- 3 {x}^{2} + 12 x - 9 = 0$
$\left(- 3 x + 3\right) \left(x - 3\right) = 0$
Hence, the factors of the equation is (-3x+3) and (x-3), solve for x,
$\left(- 3 x + 3\right) = 0$
$x = \frac{- 3}{-} 3$
$x = 1$

(x-3) =0,
x=3
Hence the x intercepts are x=1 and x =3.