The derivative of the function f is given by f'(x) = -2x + 4 for all x, and f(-1) = 5. What is the equation for the line tangent to the graph of f at x = -1.

Answer 1

integrate -2x + 4 to get -x^2 + 4x + C

set it equal to 5 for f(-1) to solve for C

f(-1)=-(-1)^2 +4(-1) + C = 5

-1-4 + C = 5

C=10

That's the graph of a downward opening parabola

f(x)=-x^2+4x+10

At the point (-1,5),the slope is f'(-1)=-2(-1)+4 = 6

Use the point slope formula

y-h = m(x-k)

y-5 = 6(x+1) or

y-6x = 11

the slope of the tangent line at x=-1 is 6 = m, the symbol for slope. (k,h) = (-1,5) the point on the graph where the tangent line just touches the graph.

Answer 2

Given: f'(x) = -2x + 4

The derivative gives the slope

of tangent line at any point.

Let m = the slope of tangent line

at point (x,y) = (-1,5)

m = f'(-1)

m = -2(-1) + 4

m = 6

The equation of a straight line is:

y = mx + b

Here: y = 6x + b

To find b (the y-intercept)

substitute (x,y) = (-1,5)

y = 6x + b

5 = 6(-1) + b

11 = b

Therefore:

y = 6x + 11

Equation of tangent line

at point (-1,5)