Vertex of a function y=(x+2)^2-9 Example-2

y=(x+2)2-9 $y=(x+2)^2-9$ , find Vertex of a function

Solution:

y=(x+2)2-9 $y=(x+2)^2-9$

1. Vertex :

$:. y=1(x-(-2))^2+(-9)$

Now compare with

$y=a(x-h)^2+k$ , we get

$a=1,h=-2,k=-9$

Vertex

$=(h,k)=(-2,-9)$

If

$a<0$ then the vertex is a maximum value

If

$a>0$ then the vertex is a minimum value

Here

$a=1>0$

So minimum Vertex =

$(h,k)=(-2,-9)$

2. Graph :
some extra points to plot the graph

$y=f(x)=(x+2)^2-9$

$f(-6)=(-6+2)^2-9=16-9=7$

$f(-5)=(-5+2)^2-9=9-9=0$

$f(-4)=(-4+2)^2-9=4-9=-5$

$f(-3)=(-3+2)^2-9=1-9=-8$

$f(-2)=(-2+2)^2-9=0-9=-9$

$f(-1)=(-1+2)^2-9=1-9=-8$

$f(0)=(0+2)^2-9=4-9=-5$

$f(1)=(1+2)^2-9=9-9=0$

$f(2)=(2+2)^2-9=16-9=7$

graph

This material is intended as a summary. Use your textbook for detail explanation.
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