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

$$12x − y + 9 = 0$$

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

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

\begin{align} y&=12x+16 \\ y&=12x-16 \\ \end{align}

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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 $y=12x-16$.

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  • $\begingroup$ where is the 8 coming from? $\endgroup$ – james Oct 2 '17 at 4:16
  • $\begingroup$ Fron $f(2)=2^3=8$. $\endgroup$ – Twink Oct 2 '17 at 16:36

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