I am first going to try finding the tangent line at this point:Take the derivative:dy/dx = (3x^2 - e^x)/2ySolve using (0, 1)dy/dx = (0 - 1)/6 = -1/6I believe that -1/6 is the slope of the tangent line, so the slope of the normal line should be +6, right?

Yep, you're correct!

[tex]e^x-x^3+y^2=10[/tex]

[tex]y^2=x^3-e^x+10[/tex]

Take the derivative of both sides.

[tex]2y \dfrac{dy}{dx} = 3x^2-e^x [/tex]

[tex]\dfrac{dy}{dx}= \dfrac{3x^2-e^x}{2y}[/tex]

Plug in (0, 3) and you get -1/6.

The derivative of a function at a certain point is the slope of the tangent of that point.

This is because the derivative is supposed to represent the rate of change for the function, and the slope of a line represents a constant rate of change. The rate of change of a function at a certain point is constant, so a line can be used to represent it. This line is the tangent line.

It is a property of perpendicular lines to have slopes that are negative reciprocals of another. The negative reciprocal of -1/6 is 6, so you are correct.