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April 23, 2014

April 23, 2014

Posted by **erika** on Monday, February 28, 2011 at 11:49pm.

- calculus -
**MathMate**, Tuesday, March 1, 2011 at 11:16amStart 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

- calculus -
**shavneel**, Saturday, June 18, 2011 at 8:38pmfind 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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