![\bf f(x)=(x-6)e^{-3x}\\\\ -----------------------------\\\\ \cfrac{dy}{dx}=1\cdot e^{-3x}+(x-6)-3e^{-3x}\implies \cfrac{dy}{dx}=e^{-3x}[1-3(x-6)] \\\\\\ \cfrac{dy}{dx}=e^{-3x}(19-3x)\implies \cfrac{dy}{dx}=\cfrac{19-3x}{e^{3x}}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3D%28x-6%29e%5E%7B-3x%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3D1%5Ccdot%20e%5E%7B-3x%7D%2B%28x-6%29-3e%5E%7B-3x%7D%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3De%5E%7B-3x%7D%5B1-3%28x-6%29%5D%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bdy%7D%7Bdx%7D%3De%5E%7B-3x%7D%2819-3x%29%5Cimplies%20%5Ccfrac%7Bdy%7D%7Bdx%7D%3D%5Ccfrac%7B19-3x%7D%7Be%5E%7B3x%7D%7D)
set the derivative to 0, solve for "x" to get any critical points
keep in mind, setting the denominator to 0, also gives us critical points, however, in this case, the denominator will never be 0, so... no critical points from there
there's only 1 critical point anyway, and do a first-derivative test on it, check a number before it and after it, to see what sign the derivative has, and thus, whether the graph is going up or down, to check for any extrema
Answer:
y=3x+1, or the second option
Step-by-step explanation:
We can see on the graph the line converges with the y-axis at (0, 1), so we can cross out the last two answers. Then we can also see that the slope is 3 on the graph and an equation that has x^2 must be a parabola. The answer is then narrowed down to the second option.