# What is the vertex of the graph of y = 2(x – 3)^2 + 4?

Jul 2, 2018

Vertex is $\left(3 , 4\right)$

#### Explanation:

If the equation of parabola is of the form $y = a {\left(x - h\right)}^{2} + k$,

the vertex is $\left(h , k\right)$.

Observe that when $x = h$, the value of $y$ is $k$ and as $x$ moves on either side, we have ${\left(x - h\right)}^{2} > 0$ and $y$ rises.

Hence, we have a minima at $\left(h , k\right)$. It would be maxima if $a < 0$

Here we have $y = 2 {\left(x - 3\right)}^{2} + 4$, hence we have vertex at $\left(3 , 4\right)$, where we have a minima.

graph{2(x-3)^2+4 [-6.58, 13.42, 0, 10]}

Jul 2, 2018

$\text{vertex } = \left(3 , 4\right)$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

•color(white)(x)y=a(x-h)^2+k

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$y = 2 {\left(x - 3\right)}^{2} + 4 \text{ is in this form}$

$\text{with "(h,k)=(3,4)larrcolor(magenta)"vertex}$

$\text{and } a = 2$

$\text{since "a>0" then graph is a minimum}$
graph{2(x-3)^2+4 [-20, 20, -10, 10]}