Question

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

Answer 1

The vertex is at:`(2, 1)`
There are no x-intercepts
The y-intercepts is at:`(0, 13)`

Explanation:

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

To find the x-intercepts set y to zero and solve for x:
`3(x-2)^2+1=0`
`3(x-2)^2=-1`
`(x-2)^2=-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(0-2)^2+1`
`y=3(-2)^2+1`
`y=3(4)+1`
`y=12+1`
`y=13`
Hence the y-intercepts is at:`(0, 13)`