The tangent to [tex]y[/tex] at the point (0, 5) has a slope equal to the derivative evaluated at [tex]x=0[/tex].
[tex]y=x^4+5e^x\implies y'=4x^3+5e^x\implies y'(0)=5[/tex]
This tangent passes through the point (0, 5), so its equation is
[tex]y-5=5(x-0)\implies y=5x+5[/tex]
The normal line passes through the same point, but is perpendicular to the tangent line, so its slope is the negative reciprocal of the slope of the tangent. So the equation for the normal line is
[tex]y-5=-\dfrac15(x-0)\implies y=5-\dfrac x5[/tex]
