If the tangent to the curve `y=x^3+a x+b` at `(1,\ -6)` is parallel to the line `x-y+5=0` , find `a` and `b` .

Answer

`Rightarrow a=-2, b=-5`

Explanation

Given: ` y=x^{3} + ax + b`........(i)

Tangent to the curve `=\frac{d y}{d x}=3 x^{2}+a`


Tangent to the curve at point `(1,-6)=\frac{d y}{d x}(1,-6)=3(1)^{2}+a=3+a`


Now, `x-y+5=0`

`\Rightarrow y=x+5`

Comparing the equation with `y=m x+c`, where `m=` slope, then we get `m=1`

Therefore, slope of the above line is `1 .`

As tangent is parallel to the line `x-y+5=0`.


...