Posted by erika on Monday, February 28, 2011 at 11:49pm.
Start with defining the function:
f(x)=ax²+bx+c
Find its derivative:
f'(x)=2ax+b
1. f'(1)=8 =>
2a(1)+b = 8 ....(1)
2. f'(-2)=-4 =>
2a(-2)+b = -4 ...(2)
Solve system of equations (1) and (2) for a and b. (a=2,b=4)
The parabola is tangent to the line y=-8 => y=-8 is the minimum.
Now find the minimum of f(x) by setting f'(x)=0 and solve for x.
f'(x)=2ax+b=0
2(2)x+(4)=0
x=-1
So
f(-1)=-8
2(-1)²+4(-1)+c = -8
so c=-6
=>
f(x) = 2x²+4x-6
Do some checking to make sure the function satisfies all the conditions (and in case I make an arithmetic error):
given conditions to be checked:
f(1)=8
f(-2)=-4
minimum f(x)=-8
find the slope of the tangent line to the graph of e^xy=x at (1,0)
write an equation of the tangent line
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