# How do you find the slope of the tangent line to the graph of the function f(t)=3t-t^2 at (0,0)?

Dec 3, 2017

$y = 3 x$

#### Explanation:

To find tangent at a point on curve $y = f \left(x\right)$, where $x = {x}_{0}$, we have the point on the curve, which is $\left({x}_{0} , f \left({x}_{0}\right)\right)$ and the slope of the tangent at that point, which is given by value of $\frac{\mathrm{df}}{\mathrm{dx}}$ at $x = {x}_{0}$.

The difference here is thatv we have $f \left(t\right) = 3 t - {t}^{2}$, i.e. $f \left(t\right)$ a function of $t$ in place of $x$. Note that $f \left(0\right) = 0$ and hence $\left(0 , 0\right)$ is on te curve.

The slope of line $f \left(t\right) = 3 t - {t}^{2}$ is given by differential of $f \left(t\right)$ at that point. As $\frac{\mathrm{df}}{\mathrm{dt}} = 3 - 2 t$ and at $t = 0$, $\frac{\mathrm{dF}}{\mathrm{dt}} = 3$

Hence, the equation of line is $\left(y - 0\right) = 3 \left(t - 0\right)$ or $y = 3 t$.

graph{(3x-x^2-y)(y-3x)=0 [-5, 5, -2.32, 2.68]}