# how to find tangent line at a given point, without equation

Find the equation of the line that is tangent to the curve at the point $(0,\sqrt{\frac{\pi}{2}})$. Given your answer in slope-intercept form.

I don't know how can I get the tangent line, without a given equation!!, this is part of cal1 classes.

• Do you mean $\frac{\sqrt{\pi}}{2}$ or $\sqrt{\frac{\pi}{2}}$? You could just respond with "the first one" or "the second one" and I will edit your question. Apr 8 '15 at 15:50
• Do you have anything? Like a picture, or a verbal description of the curve? Apr 8 '15 at 15:51
• the second one , thank you :) Apr 8 '15 at 15:52
• no, that's the problem the question kind of missing something, Apr 8 '15 at 16:00
• What is the dimension of the space where the curve lays? Apr 8 '15 at 16:00

If we suppose that your curve is the graph of a function $y=f(x)$ such that $f(0) = \sqrt{\pi/2}$, than the equation of the tangent at $x=0$ is:
$y-\sqrt{\pi/2}=f'(0)(x-0)$
$y=f'(0) x+\sqrt{\pi/2}$
So far we can infer $$T(x) = m x + n$$ with $T(0) = \sqrt{\pi/2}$. Thus $$T(x) = m x + \sqrt{\pi/2}$$ To determine the slope $m$ we need more information about the given curve.