How do you graph $y=(x+3)^2+4$ ?

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Answer 1 · 2018-05-05T13:03:54

See below.

This equation is in vertex form, where $(h,k)$ is the vertex.

Hence,

Vertex ( -3 , 4 )

When $x=0$ ,

$y=3^2+4$
$color(white)(y)=13$

Y-intercept ( 0 , 13 )

When $x=-6$ ( choose any point you want )

$y=(-3)^2+4$
$color(white)(y)=13$

( -6 , 13 )

Plot these three points and connect with a smooth curve and you should get something like this:

Graph:

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

How do you graph y=(x+3)^2+4 y=(x+3)^2+4?