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

Question

Emma Aldridge · Accepted Answer

#y=3x#

To find tangent at a point on curve #y=f(x)#, where #x=x_0#, we have the point on the curve, which is #(x_0,f(x_0))# and the slope of the tangent at that point, which is given by value of #(df)/(dx)# at #x=x_0#.

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

The slope of line #f(t)=3t-t^2# is given by differential of #f(t)# at that point. As #(df)/(dt)=3-2t# and at #t=0#, #(dF)/(dt)=3#

Hence, the equation of line is #(y-0)=3(t-0)# or #y=3t#.

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

Elijah Abram · Answer

undefined To find the slope of the tangent line to the graph of the function $ f(t) = 3t - t^2 $ at the point $(0,0)$, you need to find the derivative of the function with respect to $ t $ and then evaluate it at $ t = 0 $. The derivative of $ f(t) $ is $ f'(t) = 3 - 2t $. Evaluating $ f'(t) $ at $ t = 0 $ gives the slope of the tangent line at the point $(0,0)$. So, $ f'(0) = 3 - 2(0) = 3 $. Therefore, the slope of the tangent line to the graph of $ f(t) $ at $(0,0)$ is $ 3 $.