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

Jun 11, 2017

Vertex is at $\left(3 , - 6\right)$, y-intercept is at $\left(0 , 3\right)$
x-intercepts are at $\left(3 + \sqrt{6} , 0\right) \mathmr{and} \left(3 - \sqrt{6} , 0\right)$

#### Explanation:

$f \left(x\right) = {x}^{2} - 6 x + 3 = {\left(x - 3\right)}^{2} - 9 + 3 = {\left(x - 3\right)}^{2} - 6$. Comparing with standard equation y=a(x-h)^2+k ; (h,k) being vertex,we find here $h = 3 , k = - 6$.

So vertex is at $\left(3 , - 6\right)$

y-intercept can be found by plug in $x = 0$ in the equation.

$f \left(x\right) = 0 - 0 + 3 = 3$. So y-intercept is at $\left(0 , 3\right)$

x-intercept can be found by plug in $f \left(x\right) = 0$ in the equation.

${\left(x - 3\right)}^{2} - 6 = 0 \mathmr{and} {\left(x - 3\right)}^{2} = 6 \mathmr{and} \left(x - 3\right) = \pm \sqrt{6} \mathmr{and} x = 3 \pm \sqrt{6}$

So x-intercepts are at $\left(3 + \sqrt{6} , 0\right) \mathmr{and} \left(3 - \sqrt{6} , 0\right)$

graph{x^2-6x+3 [-40, 40, -20, 20]} [Ans]