Question 335917
How do I find the vertex and y-intercept of 

y = x² + 8x + 11, 

and how do I write it in standard form?
<pre><b>

y = x² + 8x + 11

To the side, multiply the 8 by {{{1/2}}}, getting 4
Square the 4, getting 16
Add 16 to both sides of the equation:

y + 16 = x² + 8x + 11 + 16 

Don't add the 11 and the 16 on the right, instead swap them:

y + 16 = x² + 8x + 16 + 11

Factor the first three terms on the left 

y + 16 = (x + 4)(x + 4) + 11

Write the factorization as a square:

y + 16 = (x + 4)² + 11 

Add -16 to both sides:

     y = (x + 4)² - 5

Stick a 1 in front of the parentheses:

     y = 1(x + 4)² - 5

That's the standard form. Compare to

     y = a(x - h)² + k

where (h,k) is the vertex.  So

h=-4 and k=-5 and we have that

(h,k) = (-4,-5) is the vertex.

The y-intercept is found by substituting
0 for x:

     y = (x + 4)² - 5
     y = (0 + 4)² - 5
     y = 4² - 5
     y = 16 - 5
     y = 11

So the y-intercept is (0,11)

{{{drawing(400,400,-10,10,-7,13,
locate(-8,-5,Vertex(-4,-5)),
graph(400,400,-10,10,-7,13,x^2+8x+11),
locate(0.5,11.5,y-intercept(0,11)) 


)}}}     

Edwin</pre>