SOLUTION: Given the equation y=x^2 + 16x + 8 I have to find the vertex and x intercept of this equation. I am using x = -b/2a. In trying to complete the square my answers are x= -.5 an

Algebra ->  Graphs -> SOLUTION: Given the equation y=x^2 + 16x + 8 I have to find the vertex and x intercept of this equation. I am using x = -b/2a. In trying to complete the square my answers are x= -.5 an      Log On


   

Question 753661: Given the equation y=x^2 + 16x + 8
I have to find the vertex and x intercept of this equation. I am using
x = -b/2a. In trying to complete the square my answers are x= -.5 and x= -15,5. Thus I have the x intercepts as (-.5,0) and (-15.5,0). I don't think those answers are right and I am confused about how to find the vertex.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
The vertex form of a parabola's equation is generally expressed as :
y=+a%28x-h%29%5E2%2Bk+ where (h,k) is the vertex
write your equation in vertex form
y=x%5E2+%2B+16x+%2B+8..complete the square
y=%28x%5E2+%2B+16x%2B__%29+%2B+8.....add 64 (because 8%5E2=64and 2%2A8=16) and subtract 64
y=%28x%5E2+%2B+16x%2B64%29-64+%2B+8
y=%28x+%2B+8%29%5E2-56...as you can see h=-8 and k=-56
so, the vertex is at (h,k)=(-8,-56)
x-intercept: to find them, set y=0 and solve for x
0=%28x+%2B+8%29%5E2-56
56=%28x+%2B+8%29%5E2
sqrt%2856%29=x+%2B+8
sqrt%2856%29-8=x+
solutions:
7.48-8=x+
-0.52=x+
or
-7.48-8=x+
-15.48=x+