Question

How do you find the equation of a line tangent to the function `y=x^3+6` at (1,7)?

Answer 1

`y=3x+4`

Explanation:

The tangent equation is

`y-f(x_0)=f^'(x_0)(x-x_0)`

Then `f(x_0)=7, f^'(x)=3x^2, f'(x_0)=f^'(1)=3` so the tangent is

`y-7=3(x-1)`

`y=3x+4`