# Calculus - Derivatives and Tangent Lines

I am trying to find the equation of the lines (there's 2) tangent to the graph of $f(x) = x^3$ and parallel to the line $12x − y + 1 = 0$.

So far obviously the slope of the line must be $12$. The equation of the line tangent to $f(x) = x^3$ is the derivative and must be equal to the slope of the intersecting line, $12$. The derivative of $x^3$ is equal to $2x^2$, so $2x^2 = 12$, and $x=$ $\pm\sqrt{6}$, so $y$ must equal to $\pm6\sqrt{6}$ and thus the equation of the line is $y=12x-6\sqrt{6}$ or $y=12x+6\sqrt{6}$. Is this correct? Thanks.

EDIT

I think I made a mistake it should be $y=12x-5\sqrt{6}$ or $y=12x+5\sqrt{6}$

EDIT again

Thanks to everyone who helped out! The derivative is equal to $3x^2$ not $2x^2$, so $x =\pm 2$ and so $y =\pm 8$ so plugging back into the equation gives $y=12x-16$ or $y=12x+16$, right?

• check that derivative again..... – imranfat Sep 8 '14 at 15:20
• Note that $(x^3)'\not=2x^2$. – mathlove Sep 8 '14 at 15:21
• $$y=x^3 \\y'=3x^2\\3x^2=12\\x^2=4\\x=+2,-2$$ – Khosrotash Sep 8 '14 at 15:22
• Sorry about that, it should be equal to 3x^2 right? – MathisFun Sep 8 '14 at 15:22
• @MathisFun: Yes, exactly. – mathlove Sep 8 '14 at 15:23

## 1 Answer

Yes, the slope of our line(s) needs to be $12$ to ensure they are parallel to the line $\,12x − y + 1 = 0.$

Next, finding the derivative of $y = x^3$ gives us $\,y' = 3x^2$.

To find the points on the curve at which the tangent lines at those points have slope $12$, we solve $$3x^2 = 12 \iff x^2 = 4 \iff x =\pm 2.$$

Following your edits: Yes, you are correct.

There are two lines tangent to the curve $y = x^3$ with slope $12$. One point at which a line with slope $12$ is tangent is given by $(-2, -8)$ and the other point at which a line with slope $12$ is tangent to the curve is given by $(2, 8)$.

So you have slope $m = 12$, and a point $(x_0, y_0)$ for each of two such tangent lines.

$$(y - y_0) = m(x - x_0)$$

Hence, one line is given by $$y + 8 = 12(x + 2) \iff y = 12 x + 16$$

The other line is given by $$y - 8 = 12(x -2) \iff y = 12x - 16$$