Question

How do you graph `y= - 1/2 * x^2`?

Answer 1

Refer to the explanation.

Explanation:

`y=-1/2x^2` is a quadratic equation in standard form:

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

where:

`a=-1/2`, `b=0`, and `c=0`.

You need the vertex, which is the maximum or minimum point of the parabola, and other points. Hint, since `a<0`, the vertex will be the maximum point and the parabola will open downwards.

The axis of symmetry is the `x`-value of the vertex. For a quadratic equation in standard form, the formula for determining the axis of symmetry is `x=(-b)/(2*a)`.

`x=(-0)/(-2 1/2)=0`

To find the `y`-value for the vertex, substitute `0` for the value of `x` and solve for `y`.

`y=-1/2(0)^2=0`

Vertex: `(0,0)`

Since the vertex is at the origin, and the parabola opens downward, the parabola will not cross the x- or y- axes. So we will need to choose values for `x` and solve for `y` to determine several points.

`x=1,` `y=-1/2(1)^2=-1/2`

Point: `(1,-1/2)`

`x=-1,` `y=-1/2(-1)^2=-1/2`

Point: `(-1,-1/2)`

`x=2,` `y=-1/2(2)^2=-1/2(4)=-2`

Point: `(2,-2)`

`x=-2,` `y=-1/2(-2)^2=-1/2(4)=-2`

Point: `(-2,-2)`

`x=4,` `y=-1/2(4)^2=-1/2(16)=-8`

Point: `(4,-8)`

`x=-4,` `y=-1/2(-4)^2=-1/2(16)=-8`

Point: `(-4,-8)`

Plot the points and sketch a parabola through them. Do not connect the dots.

graph{y=-1/2x^2 [-10, 10, -5, 5]}