Question

How do you find the equation of the tangent to the curve `y=x^2+2x-5` that is parallel to the line `y=4x-1`?

Answer 1

The tangent line is `y=4x+3`

To calculate this you have to follow these steps:

1) The tangent line is parallel to `y=4x+1`, so it has a form of `y=4x+c`. (1)
2) You have to find the point where the line and the function meet. To find `x` you have to calculate the 1st derivative and check where it equals 4:

`f'(x)=2x+2`
`2x+2=4`
`2x=2`
`x=1`

`f(1)=1^2+2*1-5=-2`

3) Now you have to calculate `c` for which the line (1) passes through (1,-2)

`-2=4*1+c`
`c=-6`

So finally the tangent line is: `y=4x-6`