Recall the point-slope formula:
y-y_1=m(x-x_1)
Where m=dy/dx
When x=1, we can plug that into the original equation for y
y=x^3-x^2e^(2x)
y=(1)^3-(1)^2e^(2*1)
y=1-e^2~~-6.3891
Finding the derivative of the equation for y, we get
dy/dx=d/dx(x^3)-d/dx(-x^2e^(2x))
dy/dx=3x^2-(-x^2 d/dx(e^(2x))+d/dx(-x^2)e^(2x))
dy/dx=3x^2+x^2(2e^(2x))+2xe^(2x)
When x=1, the slope of the tangent line is
dy/dx=3(1)^2+(1)^2(2e^(2(1)))+2(1)e^(2(1))
dy/dx=m=3+2e^(2)+2e^(2)=3+4e^2~~32.5562
Plugging slope, m=32.5562, the given x=1, and the derived y=-6.3891, all into the point slope formula, we get
y-y_1=m(x-x_1)
y+6.3891=32.5562(x-1)
y=32.5562x-32.5562-6.3891
y=32.5562x-38.9453
Or, if you need to be exact
y-1+e^2=(3+4e^2)(x-1)
y=3x-3+4xe^2-4e^2+1-e^2
y=(3+4e^2)x-(2+5e^2)
