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Question 805950: Find the value of a, given that the maximum value of f(x)=ax^2 +3x-4 is 5.
I'm very confused on how to do this problem. Please help!
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the value of a, given that the maximum value of f(x)=ax^2 +3x-4 is 5.
This equation is that of a parabola with a maxmum value of 5(parabola opens down)
Its standard(vertex) form of equation: y=-a(x-h)^2+k, (h,k)=(x,y) coordinates of the vertex, with k=maximum.
..
convert given equation to vertex form:
Complete the square:
y=ax^2 +3x-4
y=a(x^2+3x/a+(9/4a^2)-9/4a-4
y=a(x+3x/2a)^2-9/4a-4
(-9/4a-4)=k=5=maximum
-9/4a-4=5
-9/4a=9
-1/4a=1
4a=-1
a=-1/4
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