Find the slope and equation of tangent line at the given point.f(x)=7x-2x^2 (-1,-9)

Question

Can't figure out how to start the problem

Answer 1

To find the slope you need to differentiate the equation:

f(X)=7X-2X^2

dY/dX = 7 - 2x2X^1 = 7 - 4X

Now find what dY/dX equals at the point (-1,-9):

So, X=-1, dY/dX = 7 - (4x-1) = 7 + 4 = 11

Then the equation of a tangent can be written as:

Y - b = m(X - a)

where m = dY/dX , and (a,b) is the point on the tangent, i.e. (-1, -9)

So the equation of the tangent is:

Y - (-9) = 11(X - (-1))

Y + 9 = 11(X + 1)

Y + 9 = 11X + 11

Y = 11X + 11 - 9

Y = 11X + 2

Answer 2

Find the equation of the tangent line to the curve when x has the given value.

7) f(x) = x2 + 11x - 15 ; x = 1