If we express \($x^2 + 4x + 5$\) in the form \(a(x - h)^2 + k\) , then what is \(h\) ?

tertre
Mar 11, 2017

#1**+5 **

I'm not sure what you mean by dollars, but I'll treat that as nothing.

x^2+4x+5 --> h would be the x value of the vertex

To find the vertex we need to you the formula: -b/2a

This formula is for the axis of symmetry, but the axis of symmetry is also the x value of the vertex.

Lets state our letters:

a=1

b=4

c=5

Plug and chug into the formula

-4/2(1)

-4/2=

-2. So h =-2

Now lets find a and k

a is our strech factor which is in front of our x^2.

So a is =1

Now to find our k value which is the y value of the vertex, we plug in our x value for everytime we see x

(-2)^2+4(2)+5

4+8+5

=17 so the k=17

Now lets plug into our vertex formula

a=1, h=-2 and k=17

y=a(x-h)^2+k

NOTE for our x value of our vertex which is h we flip signs when we put it into the formula

So: y=(x+2)^2+17

Hope that helps

Julius
Mar 11, 2017