A. Find the derivative function f’ for function of f . : F(x)=3/4x+1, a=-1

Question

B. Determine an equation of the line tangent to the graph of F at (a,f(a)) for the given value of a

Answer 1

First, I would find the y-coordinate that corresponds to an x-coordinate of -1.

You can do that by repacing x in your original function with -1, and solving for y.

y = 3/4x + 1

y = 3/4(-1) + 1

y = -3/4 + 1 = -3/4 + 4/4 = 1/4.

So we will be finding the equation of the tangent line at point (-1, 1/4)

Next, we can find the f'(-1), which equals the slope of the tangent line at (-1, 1/4).

f(x) = 3/4x + 1

f'(x) = 3/4 + 0

f'(x) = 3/4

So f'(-1) = 3/4. (There is no x to replace with -1)

Now we need to find the equation of the tangent line. We know that it passes through point (-1, 1/4) and has slope 3/4. So one way to find the equation is to let x = -1, y = 1/4, and m = 3/4 in "y=mx+b", solve for b, and write the equation....

y = mx + b

1/4 = 3/4(-1) + b

1/4 = -3/4 + b

+3/4 +3/4

1 = b So the equation must be: y = 3/4x + 1