Question

How do you find the vertex and axis of symmetry, and then graph the parabola given by: `f(x)=-4x^2`?

Answer 1

The axis of symmetry is `x=0`.

The vertex is `(9,0)`.

Explanation:

Given:

`f(x)=-4x^2` is a quadratic equation in standard form:

`f(x)=ax^2+bx+c`,

where:

`a=-4`, `b=0`, `c=0`.

The axis of symmetry and the `x` coordinate of the vertex can be found using the formula:

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

`x=0/(2*0)`

`x=0`

To find the `y` coordinate of the vertex, substitute `y` for `f(x)` and `0` for `x`.

`y=-4(0)^2`

`y=0`

The vertex is `(0,0)`. Since `a<0`, the vertex is the maximum point and the parabola opens downward.

There are no x- or y-intercepts. We can determine additional points by substituting values for `x` and solving for `y`.

When `x=-1/2`,

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

`y=-4xx1/4`

`y=-1`

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

When `x=1/2`,

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

`y=-4xx1/4`

`y=-1`

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

When `x=-1`,

`y=-4(-1)^2`

`y=-4xx1`

`y=-4`

Point: `(-1,-4)`

When `x=1`,

`y=-4(1)^2`

`y=-4xx1`

`y=-4`

Point: `(1,-4)`

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

graph{y=-4x^2 [-10, 10, -5, 5]}