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 slope-4 and tangent to the line y=-8

Answer 1

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

Answer 2

find the slope of the tangent line to the graph of e^xy=x at (1,0)

write an equation of the tangent line