find the slope of the functions graph at the given point. then find an equation for the line tangent to the…

find the slope of the functions graph at the given point. then find an equation for the line tangent to the graph there.\n$f(x)=\\sqrt{7 x+1},(5,6)$\nthe slope of the functions graph at $(5,6)$ is $\\frac{7}{12}$.\n(simplify your answer.)\nthe equation for the tangent line through $(5,6)$ is $y=\\square$
Answer
Explanation:
Step1: Use the point - slope form of a line
The point - slope form of a line is (y - y_1=m(x - x_1)), where ((x_1,y_1)=(5,6)) and (m = \frac{7}{12}).
Step2: Substitute the values into the formula
Substitute (x_1 = 5), (y_1=6) and (m=\frac{7}{12}) into (y - y_1=m(x - x_1)). We get (y - 6=\frac{7}{12}(x - 5)).
Step3: Simplify the equation
[ \begin{align*} y-6&=\frac{7}{12}x-\frac{35}{12}\ y&=\frac{7}{12}x-\frac{35}{12}+6\ y&=\frac{7}{12}x-\frac{35}{12}+\frac{72}{12}\ y&=\frac{7}{12}x+\frac{37}{12} \end{align*} ]
Answer:
(y=\frac{7}{12}x+\frac{37}{12})