`f(x) = (x + 2)(x^2 + 5), (-1,6)` Find an equation of the tangent line to the graph of `f` at the given point.

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Given: `f(x)=(x+2)(x^2+5),(-1,6)`

Find the derivative using the Quotient Rule. Then substitute in the given x value into the f'(x) equation to calculate the slope.

`f'(x)=(x+2)(2x)+(x^2+5)(1)`

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

`f'(x)=3x^2+4x+5` 

`f'(-1)=3(-1)^2+4(-1)+5`

`f'(-1)=3-4+5`

`f'(-1)=4`

Use the given point (-1, 6) and the slope 4 to write an equation of the line that is tangent to the function at the specified point.

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

`y-6=4x+4`

`y=4x+10`