How do you find the vertex and intercepts for y = 3(x - 2)^2+1?

Mar 17, 2016

The vertex is at:$\left(2 , 1\right)$
There are no x-intercepts
The y-intercepts is at:$\left(0 , 13\right)$

Explanation:

$y = a {\left(x - h\right)}^{2} + k$ => Equation of the parabola in vertex form where the vertex is at:$\left(h , k\right)$, so in this case:
$y = 3 {\left(x - 2\right)}^{2} + 1$ , the vertex is at:$\left(2 , 1\right)$

To find the x-intercepts set y to zero and solve for x:
$3 {\left(x - 2\right)}^{2} + 1 = 0$
$3 {\left(x - 2\right)}^{2} = - 1$
${\left(x - 2\right)}^{2} = - \frac{1}{3}$ => Since the square of a number can not be negative the roots of this quadratic equation are complex i.e: no real roots, which means the parabola does not cross the x-axis hence there are no x-intercepts.

To find the y-intercept set x to zero and solve for y:
$y = 3 {\left(0 - 2\right)}^{2} + 1$
$y = 3 {\left(- 2\right)}^{2} + 1$
$y = 3 \left(4\right) + 1$
$y = 12 + 1$
$y = 13$
Hence the y-intercepts is at:$\left(0 , 13\right)$