find an equation of the line tangent to the graph of g(x)=4e^(-x) at the point (0,4). the equation of the…

Explanation:

Step1: Find the derivative of $G(x)$

Using the chain - rule, if $G(x)=4e^{-x}$, then $G^\prime(x)=4e^{-x}\times(- 1)=-4e^{-x}$.

Step2: Evaluate the derivative at $x = 0$

Substitute $x = 0$ into $G^\prime(x)$. $G^\prime(0)=-4e^{0}=-4$. The value of the derivative at $x = 0$ is the slope $m$ of the tangent line, so $m=-4$.

Step3: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(0,4)$ and $m=-4$. $y - 4=-4(x - 0)$.

Step4: Simplify the equation

$y-4=-4x$, so $y=-4x + 4$.

Answer:

$y=-4x + 4$