SOLUTION: looking for more help please, Thank you... Given f(x) = x^2 – 8x - 20 equation (parabola), find the following: 21.) The vertex 22.) The y-intercept(s)

Algebra ->  Graphs -> SOLUTION: looking for more help please, Thank you... Given f(x) = x^2 – 8x - 20 equation (parabola), find the following: 21.) The vertex 22.) The y-intercept(s)       Log On


   

Question 64480: looking for more help please, Thank you...
Given f(x) = x^2 – 8x - 20 equation (parabola), find the following:
21.) The vertex
22.) The y-intercept(s)
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
The easiest way to find the vertex of f(x)= ax^2 + bx + c is x=-b%2F%282a%29+
In this case, x+=+-%28-8%29%2F%282%2A1%29+=+4


To find the y coordinate of the vertex, "Plug it in, plug it in!":
y+=+x%5E2-8x-20
y=4%5E2+-8%2A4-20=+-36
Vertex = (4, -36)
Y intercept is always where x=0, so y = -20.
graph+%28300%2C600%2C+-4%2C+12%2C+-40%2C+40%2C+x%5E2+-8x-20%29+

R^2 at SCC