Question

How do you find the equation of the line tangent to `f(x)=x^2-2x-3` at the poin (-2,5)?

Answer 1

y = - 6x - 7

Explanation:

The equation of a line in `color(blue)"point-slope form"` is

`color(red)(|bar(ul(color(white)(a/a)color(black)(y-y_1=m(x-x_1))color(white)(a/a)|)))`
where m represents the gradient and `(x_1,y_1)" a point on the line"`

here we have (-2 ,5) a point on the line and require to find m.

For f(x) the gradient is found by obtaining f'(x) and evaluating
for x = - 2.

`f(x)=x^2-2x-3rArrf'(x)=2x-2`

and `f'(-2)=2(-2)-2=-6=m`

Using `m=-6" and " (x_1,y_1)=(-2,5)` substitute into equation.

`rArry-5=-6(x+2)rArry-5=-6x-12`

`rArry=-6x-7" is the equation of the tangent line"`