Recall that the derivative of a function, will give you the function that represents the slope of the original function at any value of x.
In other words, we only need to take the derivative of f(x) and plug in 0 for x to find the answer to this problem
f(x) = e^3x + 7
f '(x) = 3e^3x [use the chain rule on e^3x and note that the derivative of 7 is 0]
Now we can plug in 0 to f '(x)
f '(0) = 3e^3(0) = 3e^0 = 3(1) = 3
Lastly, the problem is asking us not just for the slope at x = 0, but the equation of the tangent line with this slope. Since we know the slope, we can just find the y-intercept and then use slope-intercept form.
y = mx + b
To find the y-intercept, plug in x = 0 to the original equation,
f(0) = e^3(0) + 7 = 1 + 7 = 8
So the y intercept is (0,8)
y = mx + b
y = 3x + 8
