Question

At what points are the tangents to `y = 2x^3 + 3x^2 -12x` horizontal?

Answer 1

`(-2, 21)` and `(1, -6)`

Explanation:

The tangent line being horizontal is equivalent to it having a slope of `0`. As the slope of the tangent line at a point is the derivative of the function at that point, we are looking for the solutions to the equation

`dy/dx = 0`

Taking the derivative, and using that `d/dx x^n = nx^(n-1)`, we have

`dy/dx = d/dx(2x^3+3x^2-12x+1)`

`=6x^2+6x-12`

`=6(x-1)(x+2)`

`=0`

`:. x = 1` or `x = -2`

Thus the points at which the tangent line is horizontal are

`(-2, 21)` and `(1, -6)`

graph{2x^3+3x^2-12x+1 [-39.54, 42.66, -10.43, 30.7]}

Answer 2

The points are `(-2, 21)` and `(1, -6)`.

Explanation:

Consider the following line.

graph{y = 0x + 1 [-10, 10, -5, 5]}

What can we say about this line?

We can say that it is horizontal. Let's look at the slope.

`m = (y_2 - y_1)/(x_2 - x_1) = (1 - 1)/(0 - 5) = 0/-5 = 0`

The slope is `0`. So, the slope of any horizontal line is `0`, since any other horizontal line will be parallel to this one.

Hence, we need to find the points on the derivative where the slope of the tangent is `0`.

The derivative can be found using a combination of the sum/difference and power rules.

`y' = 6x^2 + 6x - 12`

The slope of the tangent is given by plugging in a point, `x = a`, into the derivative.

Hence, we can set `y'`, the slope, to `0` and solve for `x`.

`0 = 6x^2 + 6x - 12`

`0 = 6(x^2 + x - 2)`

`0 = (x + 2)(x - 1)`

`x = -2 and 1`

All that is left to do is determine the corresponding y-coordinates that the function passes through.

`y = 2(-2)^3 + 3(-2)^2 - 12(-2) + 1 = 2(-8) + 3(4) + 24 + 1 = 21`

AND

`y = 2(1)^3 + 3(1)^2 - 12(1) + 1 = 2 + 3 - 12 + 1 = -6`

Hence, the points where the tangent is horizontal are `(-2, 21)` and `(1, -6)`.

Hopefully this helps!