Question

Find the points on the curve y = 2x3 + 3x2 − 12x + 6 where the tangent line is horizontal.

Answer 1

N.b., this is copied from an answer I wrote for someone else.

Thinking about the graph of y, the rate of change is zero whenever there is an extremum.

First, differentiate y.

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

Next, find the zeros of y' by factoring.

$6(x^2+x-2)=0 \\ x = -2, x = 1$

Now, we substitute these x-values into the original equation to find the coordinates. 

$(-2,21), (1,-6)$

Answer 2

The points on the curve $y = 2x^3 + 3x^2 - 12x + 6$ where the tangent line is horizontal are (-2, 21) and (1, -6)

What is tangent?

"It is a straight line that touches the curve exactly at one point."

What is slope?

"It is the ratio of the amount that y increases as x increases some amount."

For given question,

We have been given an equation of a curve $y = 2x^3 + 3x^2 - 12x + 6$

We need to find the points on the curve where the tangent line is horizontal.

First we find the slope of the tangent to the given curve.

$\Rightarrow \frac{dy}{dx}=\frac{d}{dx}(2x^3 + 3x^2 - 12x + 6) \\\\\Rightarrow \frac{dy}{dx}=\frac{d}{dx} 2x^3+\frac{d}{dx}3x^2-\frac{d}{dx}12x+\frac{d}{dx}6\\\\\Rightarrow \frac{dy}{dx}=6x^2+6x-12+0\\\\\Rightarrow \frac{dy}{dx}=6x^2+6x-12$

Since the tangent is horizontal line.

$\Rightarrow \frac{dy}{dx}=0\\\\\Rightarrow 6x^2+6x-12=0\\\\\Rightarrow x^2+x-2=0\\\\\Rightarrow x^2+2x-x-2=0\\\\\Rightarrow x(x+2)-1(x+2)=0\\\\\Rightarrow (x+2)(x-1)=0\\\\\Rightarrow x = -2,~~x = 1$

For x = -2,

$y = 2(-2)^3 + 3(-2)^2 - 12x + 6\\\\y=21$

For x = 1,

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

Therefore, the points on the curve $y = 2x^3 + 3x^2 - 12x + 6$ where the tangent line is horizontal are (-2, 21) and (1, -6)

Learn more about the tangent of the curve here:

https://brainly.com/question/19532911

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