I obtained $12x-22$ as among my answers however the mechanism says the wrong, you re welcome help. How have the right to I do this correctly? The slope of the right line is of course $12$.

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The tangent come the cubic will be parallel as soon as the slopes match, so we need $12 = f"(x) = 3x^2$, thus $x^2=4$, providing two areas on the curve in ~ $(-2,-8)$ and also $(2,8)$ The tangents v these 2 points cut the $y$-axis in ~ $-8+24 = 16$ and $8-24=-16$ respectively, giving

\beginaligny&=12x+16 \\y&=12x-16 \\\endalign The steep of the line $12x-y+9=0$ is $12$, for this reason the line you"re trying to find must have slope $12$. The derivative offers the slope of the tangent line to the graph of $f$, so we must have $f"(x)=3x^2=12$. Resolving for $x$ this provides $x=\pm 2$. You want an equation because that one line, so taking $x=2$, we have $f(x)=8$, for this reason the line passes through $(2,8)$ and has slope $12$. Utilizing the point-slope equation for a line, we acquire that the equation is $y=12x-16$. Thanks because that contributing response to naipublishers.comematics ridge Exchange!

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