Question

How do you find the equation of a line tangent to `y=2x^2-6x+3` at x=2?

Answer 1

THe equation is `y=2x-5`

Explanation:

Let `f(x)=2x^2-6x+3`
Then `f'(x)=4x-6`
`:.` `f(2)=2*4-6*2+3=-1`
`f'(2)=4*2-6=2`
The slope of the line is `=2`
So the equation of the line is `y+1=2(x-2)`
Simplifying we get, `y=2x-5`

graph{((y-2x^2-6x+3)(y-2x+5))=0 [-11.37, 8.63, -11.2, -1.2]}