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

1 Answer
Sep 4, 2016

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:

grapher

Hopefully this helps!