Question

find the slope of the functions graph at the given point. then find an equation for the line tangent to the graph there.
$f(x)=\sqrt{7 x+1},(5,6)$
the slope of the functions graph at $(5,6)$ is $\frac{7}{12}$.
(simplify your answer.)
the equation for the tangent line through $(5,6)$ is $y=\square$

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

$$
LATEXBLOCK0
$$

Answer:

\(y=\frac{7}{12}x+\frac{37}{12}\)