Answer:
Step-by-step explanation:
We will first find the derivative of the function to get the formula for the slope of the tangent. Using implicit differentiation, we find that:
[tex]6y\frac{dy}{dx}-4y\frac{dy}{dx}=-4x^3[/tex] and
[tex]\frac{dy}{dx}(6y-4y)=-4x^3[/tex] Solving for the derivative and at the same time simplifying within the parenthesis:
[tex]\frac{dy}{dx}=\frac{-4x^3}{2y}[/tex] so
[tex]\frac{dy}{dx}=\frac{-2x^3}{y}[/tex]
If our tangent line is vetical, that means that the slope is undefined. Our slope is undefined where y = 0. Therefore, we need to find x when y = 0 in our original equation. If y = 0, then
[tex]y^6-y^4=-x^4[/tex] becomes
[tex]0^6-0^4=-x^4[/tex] and
[tex]0 = x^4[/tex]
That means that x = 0 (we divide away the negative on the x, and since 0 isn't negative or positive, we get that x^4 = 0)
That means that the point at which the tangent is vertical is (0, 0)
