Question

The tangent to the curve at `y=x^2+ax+b` where a and b are constants is `2x+y=6` at the point where x=1 and how do you find a and b?

Answer 1

`a=-4, b=7 -> y = x^2-4x+7`

Explanation:

`y=x^2+ax+b`

`y' =2x+a`

We are told that the tangent at `x=1` is `2x+y=6`
`-> y=-2x+6`
Therefore the slope of the tangent `=-2`

Since the slope of y at `x=1` is given by `y'(1)`
`2(1) + a = -2`
`a=-4`

Now notice that `y(1) = 1^2+(-4)*1 +b`
`y(1)= -3+b`

Since the tangent touches `y` at `x=1`
`y(1)=-2(1)+6 -> y(1)=4`

Therefore: `-3+b =4 -> b=7`

Hence: `a=-4, b=7`
`y = x^2-4x+7`