2where is the vertex 2y+x^2=3?

Answer 1

You first have to solve for y in terms of x:

2y = 3-x^2
y = 3/2 - 1/2x^2

Now just use -b/2a for the vertex x value and then take that and plug it into the eqn for the y-value.

Answer 2

You can do this using common sense or calculus. I am going to assume you don't know calculus so this is how you do the question:

The above relation is a parabola. Knowing that a parabola is symmetric shape you can visualize that the x-value of the vertex will be the average of the x-value of the roots:

2(0)+x^2=3
x= +or- 3^(1/2)
the average of +or- 3^(1/2) = 0
so to find y: 2y+0^2=3
so y = 3/2

So the vertex is at (0, 3/2)

Answer 3

Rewrite the equation as y = 3/2 - x^2/2. The vertex is a maximum or minimum point. Take the derivative dy/dx (= - x) and set it to zero. This gives x = 0. Put that back into the original equation to get y=3/2, so the vertex is at (0, 3/2)