SOLUTION: Can someone please help me with this problem? I can't get it. Given f(x) = x^2 – 8x - 20 equation (parabola), find the following: 21. The vertex 22. The y-intercept(s) Th

Algebra ->  Linear-equations -> SOLUTION: Can someone please help me with this problem? I can't get it. Given f(x) = x^2 – 8x - 20 equation (parabola), find the following: 21. The vertex 22. The y-intercept(s) Th      Log On


   

Question 64803: Can someone please help me with this problem? I can't get it.
Given f(x) = x^2 – 8x - 20 equation (parabola), find the following:
21. The vertex
22. The y-intercept(s)
Thanks a lot
Found 2 solutions by Earlsdon, venugopalramana:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Find the vertex and y-intercept of:
f%28x%29+=+x%5E2+-+8x+-+20 Rewrite as:
y+=+x%5E2+-+8x+-+20
The x-coordinate of the vertex is given by:
-b%2F2a (See the general form:ax%5E2+%2B+bx+%2B+c+=+0
In your equation, a = 1 and b = -8
x+=+-%28-8%29%2F2%281%29
x+=+4 Substitute this into the original equation and solve for y.
y+=+%284%29%5E2+-+8%284%29+-+20
y+=+16+-+32+-+20
y+=+-36
The vertex is at: (4, -36)
The y-intercept is found by substituting x = 0 and solving for y.
y+=+%280%29%5E2+-+8%280%29+-+20
y+=+-20
The y-intercept is at: (0, -20)
graph%28300%2C200%2C-5%2C12%2C-38%2C5%2Cx%5E2-8x-20%29

Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Can someone please help me with this problem? I can't get it.
Given f(x) = x^2 – 8x - 20
y = [x^2-2x*4+4^2]-4^2-20
y = (x-4)^2-36
hence x=4 and y =-36 is the vertex.....(4,-36)
hence y intercept is -36.
when y=0,we get
(x-4)^2=36
x-4 = 6 or -6
x=6+4=10.............or...........-6+4=-2
10 and 2 are the x intercepts.
equation (parabola), find the following:
21. The vertex
22. The y-intercept(s)