Question

Find the equation of the tangent line to the curve y = x^3- 4x - 5 at the point (2, -5).Tangent Line Equation:

Answer 1

Let's find the derivative of y:

`\begin{gathered} y=x^3-4x-5 \\ (dy)/(dx)=3x^2-4 \end{gathered}`

Evaluate the derivative for x = 2:

`(dy)/(dx)\begin{cases} \\ x=2\end{cases}=3(2)^2-4=12-4=8`

Now, we have the slope, let's use the point-slope formula to find the equation:

`\begin{gathered} y-y1=m(x-x1) \\ _{\text{ }}where\colon \\ (x1,y1)=(2,-5) \\ m=8 \\ y+5=8(x-2) \\ y+5=8x-16 \\ y=8x-21 \end{gathered}`

Answer:

y = 8x - 21