Question

find an equation for the tangent line to the curve at the given point. then sketch the curve and the tangent line y = 8√x, (16,32)
an equation for the tangent line is
(type an equation.)

Explanation:

Step1: Find the derivative of the function

First, rewrite $y = 8\sqrt{x}=8x^{\frac{1}{2}}$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $y'=\frac{d}{dx}(8x^{\frac{1}{2}})=8\times\frac{1}{2}x^{\frac{1}{2}-1}=4x^{-\frac{1}{2}}=\frac{4}{\sqrt{x}}$.

Step2: Find the slope of the tangent line at the given point

Substitute $x = 16$ into the derivative. When $x = 16$, $y'(16)=\frac{4}{\sqrt{16}}=\frac{4}{4}=1$. So the slope $m$ of the tangent line at the point $(16,32)$ is $1$.

Step3: 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 $(x_{1},y_{1})=(16,32)$ and $m = 1$. Substituting these values, we get $y - 32=1\times(x - 16)$.

Step4: Simplify the equation

$y-32=x - 16$, which simplifies to $y=x + 16$.

Answer:

$y=x + 16$