$\text { Find an equation of the tangent line to the graph of } y = e ^ { 3 x } \text { at the point } ( 0,1 ) \text {. }$
A) $y = 3 x + 1$
B) $y = x + 1$
C) $y = 3 x - 1$
D) $y = \ln ( 3 ) x + 1$
E) $y = 4 x + 1$

Question

$\text { Find an equation of the tangent line to the graph of } y = e ^ { 3 x } \text { at the point } ( 0,1 ) \text {. }$ 
A) $y = 3 x + 1$
B) $y = x + 1$
C) $y = 3 x - 1$
D) $y = \ln ( 3 ) x + 1$
E) $y = 4 x + 1$

Quizplus · Accepted Answer

To find the equation of the tangent line to the graph of $y = e^{3x}$ at the point $(0,1)$, we first need to find the derivative of $y$ with respect to $x$, which gives us the slope of the tangent line at any point $x$. The derivative of $y = e^{3x}$ with respect to $x$ is $y' = 3e^{3x}$. Evaluating this derivative at $x = 0$ gives $y'(0) = 3e^{0} = 3$, which is the slope of the tangent line at the point $(0,1)$. The equation of a line with slope $m$ passing through a point $(x_1, y_1)$ is given by $y - y_1 = m(x - x_1)$. Substituting $m = 3$, $x_1 = 0$, and $y_1 = 1$ into this formula gives $y - 1 = 3(x - 0)$, or $y = 3x + 1$.

Find an equation of the tangent line to the graph of y=e3x at the point (0,1). \text { Find an equation of the tangent line to the graph of } y = e ^ { 3 x } \text { at the point } ( 0,1 ) \text {. } Find an equation of the tangent line to the graph of y=e3x at the point (0,1).

$\text { Find an equation of the tangent line to the graph of } y = e ^ { 3 x } \text { at the point } ( 0,1 ) \text {. }$