SOLUTION: This maths homework is killing me. f(x)=-x^2+6x+16 a) Write f(x) in the form a(x+b)^2+c where they are constants. State the max value of f(x). b) Solve f(x)=0 and sketch t

Algebra ->  Graphs -> SOLUTION: This maths homework is killing me. f(x)=-x^2+6x+16 a) Write f(x) in the form a(x+b)^2+c where they are constants. State the max value of f(x). b) Solve f(x)=0 and sketch t      Log On


   

Question 370383: This maths homework is killing me.
f(x)=-x^2+6x+16
a) Write f(x) in the form a(x+b)^2+c where they are constants. State the max value of f(x).
b) Solve f(x)=0 and sketch the curve y=f(x). What are the coordinates where the curve meets the x and the y axes? and the coordinates of the max or min points?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
This maths homework is killing me.
f(x)=-x^2+6x+16
a) Write f(x) in the form a(x+b)^2+c where they are constants. State the max value of f(x).
-x^2+6x+16 = y
-x^2+6x = y-16
-(x^2-6x+?) = y-16-?
-(x^2-6x+9) = y-16-9
-(x-3)^2 = y-25
y = -(x-3)^2 + 25
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Since the parabola opens downward, max occurs at the vertex
Vertex: (3,25)
Max value = 25
graph%28400%2C300%2C-10%2C10%2C-10%2C40%2C-x%5E2%2B6x%2B16%29
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b) Solve f(x)=0 and sketch the curve y=f(x).
Solve: -x^2+6x+16 = 0
x = [-6 +- sqrt(36-4*-1*16)]/-2
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What are the coordinates where the curve meets the x and the y axes? and the coordinates of the max or min points?
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Cheers,
Stan H.