Question

How do you find the equation of a line tangent to `y=-5x^2+4x-3` at x=1?

Answer 1

The equation of the line is `y(x) = -6x+2`

Explanation:

The equation of the line tangent to the curve `y=f(x)` for `x=barx` is:

`y(x) = f(bar x) + f'(bar x)(x- barx)`

Here:

`f(x) =-5x^2+4x-3`
`f'(x) =-10x+4`

and for `bar x =1`

`f(barx) = -4`
`f'(barx) = -6`

so the t the tangent line is:

`y(x) = -4-6(x-1)`

or:

`y(x) = -6x+2`