Question

let $f(x)=6x + 12-17e^{x}$. then the equation of the tangent line to the graph of $f(x)$ at the point $(0, - 5)$ is given by $y=$

Explanation:

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

The derivative of $y = 6x+12 - 17e^{x}$ using the sum - rule and basic derivative formulas. The derivative of $6x$ is $6$, the derivative of a constant $12$ is $0$, and the derivative of $- 17e^{x}$ is $-17e^{x}$. So $f^\prime(x)=6 - 17e^{x}$.

Step2: Evaluate the slope of the tangent line at $x = 0$

Substitute $x = 0$ into $f^\prime(x)$. We get $m=f^\prime(0)=6-17e^{0}=6 - 17=- 11$.

Step3: Use the point - slope form of a line

The point - slope form of a line is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(0,-5)$ and $m=-11$. Substituting these values, we have $y-(-5)=-11(x - 0)$.

Step4: Simplify the equation

$y + 5=-11x$, so $y=-11x - 5$.

Answer:

$y=-11x - 5$