Question

Find an equation for the line tangent to y = - 3 - 3x^2 at (5, - 78). The equation for the line tangent to y = - 3 - 3x^2 at (5, – 78)y = ?((I get really close and then I get it wrong. I got the slope as -90??))

Answer 1

Let's find the derivative of y:

`\begin{gathered} y=-3-3x^2 \\ (dy)/(dx)=-6x \end{gathered}`

With the derivative of the function we know its slope, therefore:

`(dy)/(dx)\begin{cases}x=5 \\ \end{cases}=-30`

Using the point-slope equation:

`\begin{gathered} Let \\ (x1,y1)=(5,-78) \end{gathered}`
`\begin{gathered} y-y1=m(x-x1) \\ y-(-78)=-30(x-5) \\ y+78=-30x+30 \\ y=-30x+72 \end{gathered}`