calculus
posted by erika .
find the equation of a quadratic function whose graph is tangent at x=1 to the line whose slope8, tangent at x=2 to solve the line with slope4 and tangent to the line y=8

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+4x6
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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