Question

Calculus Help

Answer 1

number 26

Answer 2

nvm

Answer 3

i need this one now

Answer 4

$$
f(x)=x^2+2x+1,<-3,4>\\
f'(x)=2x+2\\
f'(-3)=2(-3)+2=-6+2=-4\\
$$
Equation of the line tangent to f(x) at <-3,4> is:
$$
y=mx+b\\
4=-4(-3)+b\\
b=4-12=-8\
$$
So equation of the line tangent to f(x) at <-3,4> is
$$
y=-4x-8
$$
Make sense?

Answer 5

Slope of tangent to y at <-4,4) is
$$
m=\cfrac{4-6}{-1-3}=\cfrac{-2}{-4}=\cfrac{1}{2}
$$
So, equation of the line tangent to y at <-1,4> is
$$
z(x)=\cfrac{1}{2}x+b\\
4=\cfrac{1}{2}(-1)+b\\
b=4+\cfrac{1}{2}\\
=\cfrac{9}{2}\\
z(x)=\cfrac{1}{2}x+\cfrac{9}{2}\\
h(-1)=4\text{ this was given}\\
h'(-1)=z(-1)=\cfrac{1}{2}(-1)+\cfrac{9}{2}=\cfrac{8}{2}=4\\
$$
Let me know if you have any questions.

Answer 6

*1st line should have said: Slope of tangent to y at <-1,4) is