Question

Finding an Equation of a tangent Line in Exercise, find an equation of the tangent line to the graph of the function at the given point.
y = (e2x + 1)3,(0 , 8)

Answer 1

Answer:

Equation of tangent to the line will be $y=6x+8$

Step-by-step explanation:

We have given equation $y=3(e^{2x}+1)=3e^{2x}+3$

We have to find the equation of tangent passing through the point (0,8)

Slope of the line will be equal to $\frac{dy}{dx}=6e^2x+0=6e^{2x}$

At point (0,8) slope will be $\frac{dy}{dx}=6e^{2\times 0}=6$

Equation of line is given by

$y-y_1=m(x-x_1)$, here m is slope of the line which is equal to 6 here

So equation of line passing through (0,8)

$y-8=6(x-0)$

$y=6x+8$

So equation of tangent to the line will be $y=6x+8$