# How do you find the vertex of f(x)=3(x+4)^2+2?

Dec 18, 2017

Compare with the general vertex form to get that the vertex is at $\left(- 4 , 2\right)$.

#### Explanation:

That function, $f \left(x\right) = 3 {\left(x + 4\right)}^{2} + 2$ is already in the vertex form, $f \left(x\right) = a \left(x - h\right) + k$.

We could then see that $k$, the $y$-coordinate of the vertex, is $2$. However, for the $x$-coordinate, in the vertex form it should be subtracted, but in our function it is added.

No worries! Since two subtractions "combine" into an addition, we could reverse this process:

$f \left(x\right) = 3 {\left(x + 4\right)}^{2} + 2 \rightarrow f \left(x\right) = 3 {\left(x - \left(- 4\right)\right)}^{2} + 2$

Now we can compare with the vertex form to see that $h$, the $x$-coordinate of the vertex, is at $- 4$.

To conclude, the vertex is at $\left(- 4 , 2\right)$. Here's what the graph looks like:

graph{y=3(x+4)^2+2 [-10, 10, -5, 5]}