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

2. y=(x+2)2-9 $y=(x+2)^2-9$ , find Properties 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. Focus :
Find

$p$ , distance from the vertex to a focus of the parabola

$p=1/(4a)=1/(4*1)=1/4$

Focus

$=(h,k+p)=(-2,-9+1/4)=(-2,-35/4)$

3. Symmetry :
Axis of symmetry is the line that passes through the vertex and the focus

$x=h=-2$

4. Directrix :
Directrix

$y=k-p=-9-1/4=-37/4$

5. 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

6. Intercepts :
Intercept :
To find the y-intercept put x=0 in

$y=(x+2)^2-9$ , we get

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

$:.$ y-intercept is

$(0,-5)$

To find the x-intercept put y=0 in

$y=(x+2)^2-9$ , we get

$=>(x+2)^2-9=0$

$=>(x+2)^2=0+9$

$=>(x+2)^2=9$

$=>x+2=+- 3$

Now,

$x+2=3$

$=>x=3-2$

$=>x=1$

Now,

$x+2=-3$

$=>x=-3-2$

$=>x=-5$

$:.$ x-intercepts are

$(1,0)$ and

$(-5,0)$

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