Calculate Equation of Line Tangent Given the Graph f'(x)

Question

I'm given the graph of f'(x) which shoes a horizontal line from (10, -15) to (14, -15). The questions asks: Given f(12)=5, find the equation of the line tangent to the graph of f at x=12. How would I do this? I know to use y-y1=m(x-x1) but how would you calculate the slope?

Answer 1

1.  find the equation of the line, given two points.  I'm sure you can do that (find line's slope, then use point-slope form). Pput that equation into the form f'(x) = ax + b. Find the slope of the tangent line by evaluating f'(12). This will be m in step 3.

 

2. Observe the given fact that (12,5) is on the graph of f, i.e. when x=12, y=5. These are the coordinates of the point of tangency in this question., i.e. (x₁,y₁) is (12,5).

 

3. now use point-slope form to write the equation of the tangent at (12,5), using m obtained in step 1.

 

Answer 2

Since f'(x) is a horizontal line over the given interval, f(x) must be a linear function over that interval.  Have you been exposed to the antiderivative or integral in your calculus class?

 

f'(x) = -15  [given]

 

f(x) = -15x + c  [the antiderivative or integral]

 

The slope of the tangent line is the coefficient of x.  Get c from

 

f(12) = 5  [given]

 

f(12) = -15(12) + c

 

5 = -15(12) + c

 

c = 185

 

f(x) = -15x + 185

 

See that

 

f'(x) = -15  [the horizontal line for f'(x)]

 

and 

 

f(12) = 5.

 

The slope of the tangent line to f(x) is -15 everywhere between the 2 given points.