Question

Find an equation of the tangent line to the graph of the function at the given point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. $$y=x^3+x(-1,-2)$$

Short answer

The equation of the tangent line to the given function at the point (-1, -2) is $y=4x+6$.

Step-by-step solution

Compute the Derivative

Differentiate the given function $y=x^3+x$. Using the power rule, the derivative $y^{\prime }=3x^2+1$. This derivative function gives the slope of the tangent line at any point on the graph of the function.

Substitute the Given Point

Substitute the given point (-1, -2) into the derivative equation to determine the slope of the tangent at that point. Thus, the slope $m=3(-1{)}^2+1=3+1=4$.

Use the Slope-Point Form

Now use the slope-point form of the line equation, which is $y-y_1=m(x-x_1)$, where $(x_1,y_1)$ is the point and $m$ is the slope. Substituting our given point and computed slope into this equation's we get $y-(-2)=4(x-(-1))$, which simplifies to $y=4x+6$.

Confirm the Results

Verify your results using a graphing utility. The tangent line equation should match the line drawn on the graph at the point (-1, -2). Also confirm that the derivative at that point matches the slope you calculated, which is 4.