Identify the equation of the normal line to the curve $y=g(p)=2.5+3.5(4^p)$ where it crosses the $y$-axis.
So I am guessing the normal line would be the inverse of the derivative function, since it is perpendicular to the tangent line. Is this correct? If so, I got the derivative to be $g'(p)=3.5\cdot 4^{p}\ln(4)$. Is this the correct derivative form? If so, how do I take the inverse of this? Thank you!
The curve that you have $y=g(p)$ crosses the $y$-axis where $p=0$, so that happens at $y = 2.5 + 3.5 = 6$. So it crosses the $y$-axis at the point $(p,y) = (0,6)$.
The tangent line to the curve at this point has slope $g'(0)$. You have $$ g'(p) = 3.5\cdot \ln(4)4^p. $$ So $g'(0) = 3.5\ln(4)\simeq \dots$
To find this derivative recall the rule that $\frac{d}{dx}a^x = \ln(a)a^x$.
Now the slope of the normal line, call this $m$ satisfies that $$ mg'(0) = -1. $$ (This is because the product of the slopes of two lines that are orthogonal is $-1$).
So you need to solve this for $m$ to find the slope of the normal line. When you have $m$ (the slope of the normal line) then you can write down an equation of the normal line because you also have a point $(0,6)$ that the line passes through.
Remember that the normal line is just a straight line, so when you know the slope $m$ and a point $(0,6)$, then you can use the point-slope formula which tells you that the equation is $$ y - 6 = m(p - 0). $$ [In general when you have a slope $m$ and a point $(x_1, y_1)$, then the equation of the line that passes through the point with the slope is $y - y_1 = m(x - x_1)$.
I tried to do this and got that $m=-.206099\dots$