The function $f$ is given by $f(x)=10 e^{-3 x}+1$. What is the equation of the line tangent to the graph of $f$ when the slope of the tangent line is equal to $-10 ?$

Step 1 ：First, we need to find the derivative of the function f(x), which represents the slope of the tangent line at any point on the graph. The derivative of f(x) is given by f'(x).

Step 2 ：f(x) = 10e^{-3x} + 1

Step 3 ：Using the chain rule, we find the derivative f'(x):

Step 4 ：f'(x) = -30e^{-3x}

Step 5 ：Now, we need to find the value of x when the slope of the tangent line is equal to -10. We can set f'(x) equal to -10 and solve for x:

Step 6 ：-10 = -30e^{-3x}

Step 7 ：Divide both sides by -30:

Step 8 ：\(\frac{1}{3}\) = e^{-3x}

Step 9 ：Take the natural logarithm of both sides:

Step 10 ：ln(\(\frac{1}{3}\)) = -3x

Step 11 ：Divide by -3:

Step 12 ：x = \(\frac{1}{3}\) ln(3)

Step 13 ：Now that we have the x-coordinate, we can find the corresponding y-coordinate by plugging x into the original function f(x):

Step 14 ：f(\(\frac{1}{3}\) ln(3)) = 10e^{-3(\(\frac{1}{3}\) ln(3))} + 1

Step 15 ：Simplify the exponent:

Step 16 ：f(\(\frac{1}{3}\) ln(3)) = 10e^{-ln(3)} + 1

Step 17 ：Use the property of logarithms and exponentials:

Step 18 ：f(\(\frac{1}{3}\) ln(3)) = 10(\(\frac{1}{3}\)) + 1

Step 19 ：f(\(\frac{1}{3}\) ln(3)) = \(\frac{10}{3}\) + 1

Step 20 ：f(\(\frac{1}{3}\) ln(3)) = \(\frac{13}{3}\)

Step 21 ：Now we have the point (\(\frac{1}{3}\) ln(3), \(\frac{13}{3}\)) on the tangent line. We also know the slope of the tangent line is -10. Using the point-slope form of a linear equation, we can find the equation of the tangent line:

Step 22 ：y - \(\frac{13}{3}\) = -10(x - \(\frac{1}{3}\) ln(3))

Step 23 ：This is the equation of the line tangent to the graph of f when the slope of the tangent line is equal to -10: \(\boxed{y - \frac{13}{3} = -10(x - \frac{1}{3} ln(3))}\)