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

2 Answers
Oct 11, 2016

(-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]}

Oct 11, 2016

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!