Question

In Exercises 107-110, (a) use a graphing utility to find the derivative of the function at the given point, (b) find an equation of the tangent line to the graph of the function at the given point, and (c) use the utility to graph the function and its tangent line in the same viewing window. \(y=\left(t^{2}-9\right) \sqrt{t+2}, \quad(2,-10)\)

Short answer

The derivative of \(y=(t^{2}-9)\sqrt{t+2}\) is \(y' = 2t \sqrt{t+2} + \frac{t^{2} - 9}{2\sqrt{t+2}} \) and the equation of the tangent line is \(y = 5 \frac{1}{4}(t - 2) - 10\).

Step-by-step solution

Find the derivative

First, find the derivative of the function using the product rule and the chain rule. The derivative of \(y=t^{2}-9\cdot \sqrt{t+2}\) is \(y'=2t \sqrt{t+2} +(t^{2}-9)\frac{1}{2\sqrt{t+2}}\cdot 1\), which simplifies to \(y' = 2t \sqrt{t+2} + \frac{t^{2} - 9}{2\sqrt{t+2}}\) .

Calculate the derivative at the point

We calculate \(m = y'(2)\). Insert \(t = 2\) into \(y'\). This yields \(m = 2*2*\sqrt{2+2} + \frac{2^{2}-9}{2\sqrt{2+2}} = 8 - \frac{7}{4} = 5\frac{1}{4}\).

Find the equation of the tangent

Now you can calculate the tangent line using the point-slope form \(y = mx - mx1 + y1\). This yields \(y = 5 \frac{1}{4}(t - 2) - 10\).

Graph both function and tangent

Lastly, we graph both the function and its tangent using the graphing utility or plotting software of your choice.