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

tertre  Mar 11, 2017
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 #1
avatar+214 
+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

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