SOLUTION: A graph of f(x)=ax^2+bx+c and its turning point is S(1;18).P and T are x-intercepts of f.The graph of g(x)=-2x+8 has an x-intercept at T.R is a point of intersection of f and g.Wha

Algebra ->  Functions -> SOLUTION: A graph of f(x)=ax^2+bx+c and its turning point is S(1;18).P and T are x-intercepts of f.The graph of g(x)=-2x+8 has an x-intercept at T.R is a point of intersection of f and g.Wha      Log On


   

Question 965888: A graph of f(x)=ax^2+bx+c and its turning point is S(1;18).P and T are x-intercepts of f.The graph of g(x)=-2x+8 has an x-intercept at T.R is a point of intersection of f and g.What is the equation for f in the form of f(x)=ax^2+bx+c

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
f%28x%29=d%28x-1%29%5E2%2B18
So then to find P and T,
d%28x-1%29%5E2%2B18=0
d%28x-1%29%5E2=-18
%28x-1%29%5E2=-18%2Fd
x-1=0+%2B-+sqrt%28-18%2Fd%29
x=1+%2B-+sqrt%28-18%2Fd%29
Also you can find T.
-2x%2B8=0
2x=8
x=4
So then you can find T.
1%2Bsqrt%28-18%2Fd%29=4
sqrt%28-18%2Fd%29=3
-18%2Fd=9
d=-2
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f%28x%29=-2%28x-1%29%5E2%2B18
f%28x%29=-2%28x%5E2-2x%2B1%29%2B18
f%28x%29=-2x%5E2%2B4x-2%2B18
f%28x%29=+-2x%5E2%2B4x%2B16+
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