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?
Answer by Edwin McCravy(20054)   (Show Source):
You can put this solution on YOUR website! How do I find the vertex and y-intercept of
y = x² + 8x + 11,
and how do I write it in standard form?
y = x² + 8x + 11
To the side, multiply the 8 by  , 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)
Edwin
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