# What is the vertex of y= 3(x - 1)^2 + 1?

Jan 26, 2016

The vertex for this parabola is $\left(1 , 1\right)$.

#### Explanation:

This is an equation for a parabola. $y = 3 {\left(x - 1\right)}^{2} + 1$ is in vertex form, $y = a {\left(x - h\right)}^{2} + k$, where $a = 3 , x = h , k = 1$

The vertex is the minimum or maximum point on the parabola. In this case the vertex is the minimum point because $a$ is greater than one, so the parabola opens upward.

The axis of symmetry is the vertical line, $x$, that divides the parabola into two equal halves. In vertex form, the axis of symmetry is designated as $\left(x = h\right)$, so $\left(x = 1\right)$.

The vertex point in vertex form is $\left(h , k\right)$, which is $\left(1 , 1\right)$.

graph{y=3(x-1)^2+1 [-10, 10, -5, 5]}