Find the tangent line to a curve at a given value

A common early calculus problem is finding the tangent line to a curve at a given value. Consider the following problem:

Find the equation of the tangent line at x = -1 to the following curve:

f(x) = 3x³ + 7x² - 8x - 4

First you need to find the derivative in order to find the slope of the tangent line. In case you don't remember derivatives, here is how basic derivatives work:

• If f(x) = x^a, then the derivative f '(x) = ax^(a-1).
• So, you bring the exponent down as a multiplier and then subtract 1 from the exponent.
• There are some exceptions, but that is how to do derivatives in general.
• If f(x) = x, that is x¹. So f '(x) = 1x⁰, which is 1.
• If f(x) = 3, that is 3x⁰. So f '(x) = 0(3x^-1) = 0. The derivative of an actual number is 0.

Now back to the problem:

Find the tangent line at x = -1 to the curve f(x) = 3x³ + 7x² - 8x - 4

The derivative to find the slope of the tangent line:

f '(x) = 3(3x)² + 2(7x) - 8

f '(x) = 9x² + 14x - 8

Next we need to use the point-slope form to find the tangent line. Here it is:

y - y₁ = m(x - x₁)

We have to find y₁ and x₁ and the slope m.

Now, recall we want the tangent line for x = -1. Find y₁ by solving for f(x) at the given value. So, If we solve for f(x) where x = -1, we have the following:

f(-1) = 3(-1)³ + 7(-1)² - 8(-1) - 4 = -3 + 7 + 8 - 4 = 8

y₁ = 8

Since we are looking for an answer at x = -1, x₁ = -1.

And recall that the derivative is the slope. So, we solve f '(-1):

f '(-1) = 9(-1)² + 14(-1) = 9 - 14 - 8 = -13

m = -13

Finally, we plug it all into the point-slope form y - y₁ = m(x - x₁):

y - 8 = -13(x -(-1))

y - 8 = -13(x + 1)

y - 8 = -13x - 13

y - 8 + 8 = -13x - 13 + 8

y = -13x - 5

And that is the equation for the tangent line to the curve f(x) = 3x³ + 7x² - 8x - 4 at x = -1.

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