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

Explanation

Given that`y=x^3+a x+b`......(1)

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 it with` y=m x+c`, where `m= slope`, we get `m=1`

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