Question

a. find an equation of the tangent line at x = a. b. use a graphing utility to graph the curve and the tangent line on the same set of axes. y = 8√x, a = 16 a. the equation of the tangent line is y =

Explanation:

Step1: Find the derivative of the function

The function is $y = 8\sqrt{x}=8x^{\frac{1}{2}}$. Using the power - rule for differentiation $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $y^\prime=\frac{d}{dx}(8x^{\frac{1}{2}})=8\times\frac{1}{2}x^{\frac{1}{2}-1}=4x^{-\frac{1}{2}}=\frac{4}{\sqrt{x}}$.

Step2: Find the slope of the tangent line at $x = a$

Substitute $x = a = 16$ into the derivative. The slope $m=y^\prime|_{x = 16}=\frac{4}{\sqrt{16}}=\frac{4}{4}=1$.

Step3: Find the y - coordinate of the point on the curve at $x = a$

Substitute $x = 16$ into the original function $y = 8\sqrt{x}$. So $y=8\sqrt{16}=8\times4 = 32$. The point on the curve is $(16,32)$.

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

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(16,32)$ and $m = 1$.
$y-32=1\times(x - 16)$.
Simplify to get $y=x+16$.

Answer:

$y=x + 16$