Question

How do you graph the quadratic function and identify the vertex and axis of symmetry for `y=-2(x+3)^2-4`?

Answer 1

See below.

Explanation:

This quadratic function will have a vertex at `(-3, -4)`, since in vertex form `y = a(x - p)^2 + q`, the vertex is located at `(p, q)`.

The equation of the axis of symmetry (it is a vertical line) will be `x= -3`. That will always be the x-coordinate of the vertex.

To graph, it is not only helpful to know the vertex and the equation of the axis of symmetry. Intercepts, both x and y, and direction of opening is extremely important.

Let's start with the latter. The parameter `a` in `y = a(x - p)^2 + q` influences the breadth and the direction of opening of the parabola. If `a > 0`, then the parabola opens upwards. Similarly, if `a < 0`, the parabola opens down.

Here, `a = -2`, so the parabola opens down.

Now for x-intercepts. These can be obtained by setting `y = 0` and solving the resulting quadratic.

`y = -2(x + 3)^2 - 4`

`0 = -2(x + 3)^2 - 4`

`4/-2 = (x + 3)^2`

`+-sqrt(-2) = x + 3`

`x = O/`

So, there is no x-intercept. We could have figured out without algebra, because the vertex lies below the x-axis and the parabola opens downwards.

As for y-intercepts, set `x = 0` and solve.

`y = -2(x + 3)^2 - 4`

`y = -2(0 + 3)^2 - 4`

`y = -2(9) - 4`

`y = -22`

`:.` The y-intercept has coordinates of `(0, -22)`.

Let's finish by identifying the domain and range. The domain is `x in RR` and the range is `y ≤ -4`, since the point `(-3, -4)` is the maximum.

We can now graph, using a table of values and connecting the points using a smooth, curved line.

Here is what your graph should look like:

Hopefully this helps!