please help with math

Question

A curve in the xy-plane is defined by the equation x^3/3+y^2/2−3x+2y=−1/6. Which of the following statements are true?

i. At points where x=√3, the lines tangent to the curve are horizontal.

ii. At points where x=-2, the lines tangent to the curve are vertical.

iii. The line tangent to the curve at the point (1,1) has slope 2/3.

a) all of them

b) ii and iii

c) i and ii

d) i and iii

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Let C be the curve defined by x^2−y^2=1. Consider all points (x,y) on curve C where x>1 and y>0. Which of the following statements provides a justification for the concavity of the curve?

a) The curve is concave down because y'' = -x/(y^2) < 0

b) The curve is concave down because y'' = -1/(y^3) < 0

c) The curve is concave up because y'' = 1/(y^2) > 0

d) The curve is concave up because y'' = 1/(y^3) > 0

Answer 1

Using implicit differentiation

But the original equation tells us x² = y² + 1 so replacing x² with y² + 1 we get

y'' = -1/y³ and since y>0 the y'' is always negative so it is concave down.

Answer 2

For the first part, differentiate the expression with respect to x to get

x² + yy' -3 + 2y' =0 where y' is the derivative with respect to x.

Solve for y'

y' = (3 - x²)/(y + 2)

when x = √3 , x² = 3 then y' = 0/(y+2) = 0 as long as y ≠ -2

A derivative of zero implies that the tangent to the curve is horizontal, so i is true.

when x = 1 and y = 1, y' = (3 - 1)/(1 + 2) = 2/3, so iii is true.

I am not sure about ii as I get an imaginary result. Sorry.