Question

find the equation of the linr tangent to the function ay the given point. Answer has to be in slope intercept form

Answer 1

$y=\frac{1}{2}x+3$

Explanation

Sep 1

Find the first derivative of f(x)

$\begin{gathered} y=(5x+5)^{\frac{1}{2}} \\ \text{derivate applyin the chain rule} \\ y^{\prime}\text{ = }\frac{1}{2}(5x+5)^{\frac{1}{2}-\frac{2}{2}}\cdot(5) \\ y^{\prime}\text{ = }\frac{1}{2}(5x+5)^{-\frac{1}{2}}\cdot(5) \\ y^{\prime}\text{ = }\frac{5}{2}(5x+5)^{-\frac{1}{2}} \\ \text{ y'= }\frac{5}{2}(5x+5)^{-\frac{1}{2}} \end{gathered}$

Step 2

Plug x value of the indicated point into f '(x) to find the slope at x.

so

Therefore, at x = 4, the slope of the tangent line is

$\begin{gathered} \text{ y'= }\frac{5}{2}(5(4)+5)^{-\frac{1}{2}} \\ \text{ y'= }\frac{5}{2}(25)^{-\frac{1}{2}} \\ \text{ y'= }\frac{5}{2}\cdot\frac{1}{\sqrt[]{25}} \\ \text{ y'= }\frac{5}{2}\cdot\frac{1}{5}=\frac{1}{2} \\ so,\text{ the slope is 1/2} \end{gathered}$

Step 3

finally, find the equation of the lne by using

$\begin{gathered} y-y_1=m(x-x_1) \\ \text{where m is the slope} \\ \text{and ( x}_1,y_1)\text{ is a point of the line, } \\ \text{hence, replace} \\ y-5=\frac{1}{2}(x-4) \\ y-5=\frac{1}{2}x-2 \\ \text{add 5 in both sides} \\ y-5+5=\frac{1}{2}x-2+5 \\ y=\frac{1}{2}x+3 \end{gathered}$

so, the answer is

$y=\frac{1}{2}x+3$

I hope this helps you