Answer:
Equation of tangent to the line will be [tex]y=6x+8[/tex]
Step-by-step explanation:
We have given equation [tex]y=3(e^{2x}+1)=3e^{2x}+3[/tex]
We have to find the equation of tangent passing through the point (0,8)
Slope of the line will be equal to [tex]\frac{dy}{dx}=6e^2x+0=6e^{2x}[/tex]
At point (0,8) slope will be [tex]\frac{dy}{dx}=6e^{2\times 0}=6[/tex]
Equation of line is given by
[tex]y-y_1=m(x-x_1)[/tex], here m is slope of the line which is equal to 6 here
So equation of line passing through (0,8)
[tex]y-8=6(x-0)[/tex]
[tex]y=6x+8[/tex]
So equation of tangent to the line will be [tex]y=6x+8[/tex]
