Let L be the tangent line to y=tan(2x) at (pi/2,0). What is the y-intercept of L? (a) 0 (b) pi/2 (c) −pi (d) 1 (e) 2

Let $L$ be the tangent line to $y=\mathrm{tan}\left(2x\right)$ at $\left(\frac{\pi }{2},0\right)$. What is the $y$-intercept of $L$?
(a) $\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}0\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}$
(b) $\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\frac{\pi }{2}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}$
(c) $\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-\pi \phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}$
(d) $\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}1\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}$
(e) $\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}2\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}$
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Absexabbelpjl
The equation of the tangent line at the point $\left(a,f\left(a\right)\right)$ is
$L\left(x\right)={f}^{\prime }\left(a\right)\left(x-a\right)+f\left(a\right)$
In this case, $f\left(x\right)=\mathrm{tan}\left(2x\right)$ and $a=\pi /2$. Hence
$L\left(x\right)=2{\mathrm{sec}}^{2}\left(2\cdot \frac{\pi }{2}\right)\left(x-\frac{\pi }{2}\right)+\mathrm{tan}\left(2\cdot \frac{\pi }{2}\right)$
i.e. $L\left(x\right)=2\left(x-\frac{\pi }{2}\right)=2x-\pi$. Therefore, the 𝑦-intercept of $L$ is $-\pi$.
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The tangent to the curve $y=f\left(x\right)$ at the point $\left(t,f\left(t\right)\right)$ is the line
$y-f\left(t\right)={f}^{\prime }\left(t\right)\left(x-t\right)$
The $y$-intercept of this line is the point where $x=0$ and hence
$y=f\left(t\right)-t{f}^{\prime }\left(t\right)$
Can you solve it now?