# How do you find the vertex and the intercepts for f(x) = 3x^2 - 6x + 8?

Mar 28, 2017

Vertex is at $\left(1 , 5\right)$ , Y-intercept is at $\left(0 , 8\right)$

#### Explanation:

$f \left(x\right) = 3 {x}^{2} - 6 x + 8 \mathmr{and} f \left(x\right) = 3 \left({x}^{2} - 2 x\right) + 8 \mathmr{and} f \left(x\right) = 3 \left({x}^{2} - 2 x + 1\right) - 3 + 8 \mathmr{and} f \left(x\right) = 3 {\left(x - 1\right)}^{2} - + 5 \therefore$

Vertex is at $\left(1 , 5\right)$ [Comparing with standard equation $f \left(x\right) = a {\left(x - h\right)}^{2} + k , \left(h , k\right)$ is vertex]

f(x) = 0+0+8=8 ; x=0
Y-intercept is at $\left(0 , 8\right)$ [Obtained by putting $x = 0$ in the equation.]

X-intercept can be obtained by putting $y = 0$ in the equation.
$3 {x}^{2} - 6 x + 8 = 0$, Discriminant $= {b}^{2} - 4 a c = - 60$ is negative, roots are complex and parabola does not cross $x$ axis

Vertex is at $\left(1 , 5\right)$ , Y-intercept is at $\left(0 , 8\right)$ graph{3x^2-6x+8 [-40, 40, -20, 20]}[Ans]