# What are the coordinates of the vertex of the parabola whose equation is y = 3(x - 2)^2 + 5?

Apr 12, 2015

The answer is: $V \left(2 , 5\right)$.

There are two ways.

First:

we can remember the equation of the parabola, given the vertex $V \left({x}_{v} , {y}_{v}\right)$ and the amplitude $a$:

$y - {y}_{v} = a {\left(x - {x}_{v}\right)}^{2}$.

So:

$y - 5 = 3 {\left(x - 2\right)}^{2}$ has vertex: $V \left(2 , 5\right)$.

Second:

we can make the counts:

$y = 3 \left({x}^{2} - 4 x + 4\right) + 5 \Rightarrow y = 3 {x}^{2} - 12 x + 17$

and, remembering that $V \left(- \frac{b}{2 a} , - \frac{\Delta}{4 a}\right)$,

$V \left(- \frac{- 12}{2 \cdot 3} , - \frac{{12}^{2} - 4 \cdot 3 \cdot 17}{4 \cdot 3}\right) \Rightarrow V \left(2 , 5\right)$.

Apr 12, 2015

Vertex is $\left(2 , 5\right)$

Method

Use the form: ${\left(x - h\right)}^{2} = 4 a \left(y - k\right)$

This parabola has vertex at $\left(h , k\right)$
And its major axis is along the $y - \text{axis}$

In our case we have, $y = 3 {\left(x - 2\right)}^{2} + 5$

$\implies 3 {\left(x - 2\right)}^{2} = y - 5$

$\implies {\left(x - 2\right)}^{2} = \frac{1}{3} \left(y - 5\right)$

So, the vertex is $\left(2 , 5\right)$

Worthy of note
When the equation is of the form: ${\left(y - k\right)}^{2} = 4 a \left(x - h\right)$

The vertex is at $\left(h , k\right)$ and the parabola lies along the $x - \text{axis}$