To find the equation of the line tangent to the graph of the function [tex]\( G(x) = 3e^{-5x} \)[/tex] at the point [tex]\( (0, 3) \)[/tex], we need to determine two key pieces of information: the slope of the tangent line at that point and the y-intercept of the tangent line.
### Step 1: Find the Slope of the Tangent Line
The slope of the tangent line at a point on a function's graph is given by the derivative of the function evaluated at that point. We first need to find the derivative of [tex]\( G(x) \)[/tex]:
[tex]\[ G(x) = 3e^{-5x} \][/tex]
Differentiate [tex]\( G(x) \)[/tex] with respect to [tex]\( x \)[/tex]:
[tex]\[ G'(x) = \frac{d}{dx}[3e^{-5x}] \][/tex]
Using the chain rule, we get:
[tex]\[ G'(x) = 3 \cdot \frac{d}{dx}[e^{-5x}] = 3 \cdot (-5e^{-5x}) = -15e^{-5x} \][/tex]
Next, we evaluate [tex]\( G'(x) \)[/tex] at [tex]\( x = 0 \)[/tex]:
[tex]\[ G'(0) = -15e^{-5(0)} = -15e^{0} = -15 \][/tex]
So, the slope of the tangent line at the point [tex]\( (0, 3) \)[/tex] is [tex]\(-15\)[/tex].
### Step 2: Find the Y-Intercept of the Tangent Line
The point-slope form of the equation of the tangent line is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( m \)[/tex] is the slope and [tex]\( (x_1, y_1) \)[/tex] is the given point on the curve.
Given:
- The slope [tex]\( m = -15 \)[/tex]
- The point [tex]\( (x_1, y_1) = (0, 3) \)[/tex]
Substitute these values into the point-slope form:
[tex]\[ y - 3 = -15(x - 0) \][/tex]
Simplify this equation to get the slope-intercept form:
[tex]\[ y - 3 = -15x \][/tex]
Add 3 to both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ y = -15x + 3 \][/tex]
Thus, the equation of the line tangent to the graph of [tex]\( G(x) \)[/tex] at the point [tex]\( (0, 3) \)[/tex] is:
[tex]\[ y = -15x + 3 \][/tex]
