Question

11 multiple choice 1 point find the equation of the line tangent to the graph of the function at the indicated point. $y = (x^2 + 28)^{4/5}$ at $x = 2$ $y = \frac{8}{5}x + \frac{64}{5}$ $y = \frac{4}{5}x + \frac{64}{5}$ $y = \frac{8}{5}x + \frac{96}{5}$ $y = \frac{8}{5}x$

Explanation:

Step1: Find the y - coordinate at x = 2

Substitute \(x = 2\) into the function \(y=(x^{2}+28)^{\frac{4}{5}}\). First, calculate \(x^{2}+28\) when \(x = 2\): \(2^{2}+28=4 + 28=32\). Then \(y=(32)^{\frac{4}{5}}\). Since \(32 = 2^{5}\), we have \(y=(2^{5})^{\frac{4}{5}}=2^{4}=16\). So the point of tangency is \((2,16)\).

Step2: Find the derivative of the function

Use the chain rule. If \(y = u^{\frac{4}{5}}\) where \(u=x^{2}+28\), then \(\frac{dy}{du}=\frac{4}{5}u^{-\frac{1}{5}}\) and \(\frac{du}{dx}=2x\). By the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{4}{5}(x^{2}+28)^{-\frac{1}{5}}\cdot2x=\frac{8x}{5(x^{2}+28)^{\frac{1}{5}}}\).

Step3: Find the slope of the tangent line at x = 2

Substitute \(x = 2\) into the derivative. We know \(x^{2}+28 = 32\) when \(x = 2\), so \((x^{2}+28)^{\frac{1}{5}}=32^{\frac{1}{5}}=(2^{5})^{\frac{1}{5}} = 2\). Then \(\frac{dy}{dx}\big|_{x = 2}=\frac{8\times2}{5\times2}=\frac{16}{10}=\frac{8}{5}\). So the slope \(m=\frac{8}{5}\).

Step4: Use the point - slope form to find the equation of the tangent line

The point - slope form of a line is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(2,16)\) and \(m=\frac{8}{5}\). Substitute these values: \(y-16=\frac{8}{5}(x - 2)\). Expand the right - hand side: \(y-16=\frac{8}{5}x-\frac{16}{5}\). Then add 16 to both sides. 16 can be written as \(\frac{80}{5}\), so \(y=\frac{8}{5}x-\frac{16}{5}+\frac{80}{5}=\frac{8}{5}x+\frac{64}{5}\).

Answer:

\(y=\frac{8}{5}x+\frac{64}{5}\) (the first option)