For what value(s) of x does the graph have a horizontal tangent?

Question

For what value(s) of x does the graph of f(x)=6x³−18x²−144x+72 have a horizontal tangent?

I imputed the equation in desmos and thought that there did not exist a horizontal tangent, but I was wrong.

Answer 1

The equation has a horizontal tangent when the slope of the tangent is zero. The slope of the tangent line to the graph of f(x) is actually f'(x). So when f'(x) = 0 we have a horizontal tangent line (the point where it changes from increasing to decreasing or vice versa).

f(x) = 6x³ - 18x² - 144x + 72

f'(x) = 18x² - 36x - 144 = 0 when slope is horizontal

Factoring out 18

18(x² - 2x - 8) = 0

Dividing both sides by 18 and factoring further

(x - 4)(x + 2) = 0

So the horizontal tangents are at x = 4, and x = -2

Answer 2

Horizontal tangents can occur at 3 different types of critical points: a relative minimum, a relative maximum, or a flat point of inflection. The first two should be easy to spot on a graph; the third is a little less obvious. Think of how y = x³ as it passes through the origin, as an example of the a flat inflection point.

In any case to find any horizontal tangents, we simply find the derivative and set it = 0. If that equation has any solutions, those solutions are places on the graph of the parent function where the tangent line is horizontal:

f^'(x) = 18x² - 36x - 144. Now we set it = 0, and since it is a quadratic, we have simple ways to find solutions:

18x² - 36x - 144 = 0. Divide everything by 18

x² - 2x - 8 = 0

(x - 4)(x+2) = 0

x = 4 or x = -2 Because f(x) is of odd degree with a + leading coefficient, its graph starts in QIII and ends in QI.

This is one way to figure out that x = -2 is a relative max and x = 4 is a relative min.

Try graphing it again on Desmos and see if you can spot those two stationary points now that you know where to look! Happy hunting!