# Find an equation of a line that is tangent to the graph of $f$ and parallel to the given line.

$$f(x) = x^3$$

$$12x − y + 9 = 0$$

I got $12x-22$ as one of my answers but the system says its wrong, please help. How can I do this correctly?

• Can you check your equations: did you really mean $12^x$ or just $12x$? – Joffan Oct 2 '17 at 0:57
• no, i double checked and It's just 12x – james Oct 2 '17 at 1:02
• Can you change it in your question? – user222031 Oct 2 '17 at 1:10
• just did thank you so much, sorry i didnt catch that – james Oct 2 '17 at 1:11
• If you’d like someone to point out where you went wrong, then show your work. – amd Oct 2 '17 at 7:07

The slope of the straight line is of course $12$.
The tangent to the cubic will be parallel when the slopes match, so we need $12 = f'(x) = 3x^2$, thus $x^2=4$, giving two locations on the curve at $(-2,-8)$ and $(2,8)$ The tangents through these two points cut the $y$-axis at $-8+24 = 16$ and $8-24=-16$ respectively, giving
The slope of the line $12x-y+9=0$ is $12$, so the line you're looking for must have slope $12$. The derivative gives the slope of the tangent line to the graph of $f$, so we must have $f'(x)=3x^2=12$. Solving for $x$ this gives $x=\pm 2$. You want an equation for one line, so taking $x=2$, we have $f(x)=8$, so the line passes through $(2,8)$ and has slope $12$. Using the point-slope equation for a line, we get that the equation is $y=12x-16$.
• Fron $f(2)=2^3=8$. – Twink Oct 2 '17 at 16:36