Question

How do you find the points where the graph of the function `f(x) = x^3 - 4x^2 - 7x + 8` has horizontal tangents and what is the equation?

Answer 1

See below.

Explanation:

The tangent line is horizontal when the slope of the tangent line is `0`.

The slope of the tangent line is `0` when the derivative is `0`.

So to find the `x`-coordinates at which the tangent line is horizontal, we need to solve `f'(x) = 0`.

For this function,

`f(x) = x^3-4x^2-7x+8`, we get

`f'(x) = 3x^2-8x-7 = 0`

I assume a calculus student knows the quadratic formula or completing the square, so solve the equation to get

`x = (4+-sqrt37)/3`.

There are two places where the tangent line is horizontal.

At `x = (4-sqrt37)/3`, we get

`y = f((4-sqrt37)/3) = ((4-sqrt37)/3)^3-4((4-sqrt37)/3)^2-7((4-sqrt37)/3)+8`

`= (-164+74sqrt37)/27`

The tangent line is horizontal. The equation of a horizontal line through the point `(h,k)` is `y = k`, so the tangent line to the graph of `f(x) = x^3-4x^2-7x+8` at `x = (4-sqrt37)/3` is

`y = (-164+74sqrt37)/27`

At the other point we have `x = (4+sqrt37)/3`. So,

`y = f((4+sqrt37)/3) = ((4+sqrt37)/3)^3-4((4+sqrt37)/3)^2-7((4+sqrt37)/3)+8`

`= (-164-74sqrt37)/27`

And the equation of that tangent line is

`y = (-164-74sqrt37)/27`