Question

How do you tell whether the graph opens up or down, find the vertex, and find the axis of symmetry given `y=2x^2-2`?

Answer 1

Refer to the explanation.

Explanation:

Given:

`y=2x^2-2` is a quadratic equation in standard form:

`y=ax^2+bx+c`,

where:

`a=2`, `b=0`, and `c=-2`.

Axis of Symmetry: the line that divides the parabola into two equal halves.

The formula for the axis of symmetry is:

`x=(-b)/(2a)`

`x=0/(2*2)`

`x=0`

The axis of symmetry is `0`. This is also the `x` value of the vertex.

Vertex: minimum or maximum point of the parabola

`x=0`. Substitute `0` for `x` and solve for `y`.

`y=2(0)^2-2`

`y=0-2`

`y=-2`

The vertex is `(0,-2)`.

Since `a>0`, the vertex is the minimum point, and the parabola will open upward.

graph{y=2x^2-2 [-10, 10, -5, 5]}

Answer 2

Well, here's how I'd do it:

Explanation:

Since it's algebra and not calculus. Calculus gives you some better tools to more quickly answer these questions, but I won't use them.

note that for x = 0, y = -2. This is the minimum value - so it's your vertex.
Any value of x > 0 will give a value for y > -2.
Also, note that if x is negative ( < 0), `x^2` is positive, since any negative number multiplied by another negative number is positive.
So, for any value of x < 0, y will be > -2.
This tells you that your graph opens UP.

The axis of symmetry is that point a on the x axis where, if you add some number (call it b) to it and calculate y, you get the same value when you subtract that number from it and calculate y.

In other words, f(a+b) = f(a-b).

You already pretty much know that this is x = 0, but you can solve for it algebraically:

`2(a+b)^2 - 2 = 2(a-b)^2 -2` ...and we want to solve for a.

add 2 to both sides:
`2(a+b)^2 = 2(a-b)^2`

then multiply everything out...
`2(a^2 + 2ab + b^2) = 2(a^2 - 2ab + b^2)`
`2a^2 + 4ab + 2b^2 = 2a^2 -4ab + 2b^2`

then subtract `2a^2 + 2b` from both sides...
`4ab = -4ab`

divide by 4b on both sides gives you:

a = -a

which is true ONLY when a = 0. So therefore a is your axis of symmetry