# What are the important points needed to graph  y = 3( x + 1 )^2 -4?

Jun 22, 2018

see graph.

#### Explanation:

this is in vertex form:

$y = a {\left(x + h\right)}^{2} + k$

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

Axis of symmetry $a o s = - h$

$a > 0$ open up, has a minimum

$a < 0$ opens down has a maximum

you have:

vertex #(-1,-4)

$a o s = - 1$

set $x = 0$ to solve y-intercept:

$y = 3 {\left(x + 1\right)}^{2} - 4$

$y = 3 {\left(0 + 1\right)}^{2} - 4 = - 1$

$y = - 1$

set $y = 0$ to solve x-intercept(s) if they exist:

$y = 3 {\left(x + 1\right)}^{2} - 4$

$0 = 3 {\left(x + 1\right)}^{2} - 4$

$\frac{4}{3} = {\left(x + 1\right)}^{2}$

$\pm \sqrt{\frac{4}{3}} = x + 1$

$x = - 1 \pm \sqrt{\frac{4}{3}}$

$a = 5$ so $a > 0$ parabola opens up and has a minimum at vertex.

graph{3( x + 1 )^2 -4 [-10, 10, -5, 5]}