Question

find the equation of the tangent line to the graph of (f(x)) at the point ((0, 12)).

(f(x) = 10x + 15 - 3e^x)

write the answer in slope-intercept form, i.e., (y = mx + b).

(y_{\text{tan}} = \\)

Explanation:

Find the derivative of the function

$$ f'(x) = \frac{d}{dx}(10x + 15 - 3e^x) = 10 - 3e^x $$

Evaluate the derivative at the given point

$$ m = f'(0) = 10 - 3e^0 = 10 - 3(1) = 7 $$

Write the tangent line equation

$$
LATEXBLOCK0
$$

Answer:

Find the equation of the tangent line to the graph of \(f(x)\) at the point \((0, 12)\).
\(f(x) = 10x + 15 - 3e^x\)
Write the answer in slope-intercept form, i.e., \(y = mx + b\).

\(y_{\text{tan}} =\) <blank>\(7x + 12\)</blank>