# How do you find the vertex and the intercepts for y = 3x^2 - 18x + 4?

May 29, 2017

Vertex is at $\left(3 , - 23\right)$ , y-intercept is at $\left(0 , 4\right)$ and
x-intercepts are at
$\left(0.23 , 0\right) \mathmr{and} \left(5.77 , 0\right)$

#### Explanation:

$y = 3 {x}^{2} - 18 x + 4 = 3 \left({x}^{2} - 6 x\right) + 4 = 3 \left({x}^{2} - 6 x + 9\right) - 27 + 4$ or
$y = 3 {\left(x - 3\right)}^{2} - 23$ Comparing with vertex form of equation

y=a(x-h)^2+k ; (h,k) being vertex we find here  h=3 ; k=-23

So vertex is at $\left(3 , - 23\right)$ , y-intercept can be found by putting $x = 0$ in the equation :.y=4 ; y-intercept is at $\left(0 , 4\right)$

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

$\therefore 3 {\left(x - 3\right)}^{2} - 23 = 0 \mathmr{and} 3 {\left(x - 3\right)}^{2} = 23 \mathmr{and} {\left(x - 3\right)}^{2} = \frac{23}{3}$ or

$\left(x - 3\right) = \pm \sqrt{\frac{23}{3}} \mathmr{and} x = 3 \pm \sqrt{\frac{23}{3}} \therefore x \approx 5.77 \left(2 \mathrm{dp}\right) \mathmr{and} x \approx 0.23 \left(2 \mathrm{dp}\right)$

x-intercepts are at $\left(0.23 , 0\right) \mathmr{and} \left(5.77 , 0\right)$
graph{3x^2-18x+4 [-80, 80, -40, 40]} [Ans]