Identify the vertex, axis of symmetry, focus and directrix of the parabola below

Question

the equation of a parabola is y=1/4(x+3)²-1

Answer 1

To start, let's find the vertex. The equation here is given in vertex form, which is easier to work with. Vertex form for a quadratic is as follows: y=a(x-h)²+k. Which is identical to what you have. The vertex is always going to be (h,k).

So in this case, the vertex is (-3,-1).

Next let's move onto the axis of symmetry. It is always going to be along the x value if the parabola opens up or down. If it opens to the left or right, symmetry will be along the y.

In this case, the axis of symmetry is x=-3.

Now for the focus, the formula for the focus is (h,k+1/[4a]). We can just plug in what we have and solve. It will work out nicely in this one.

(-3,-1+1/[4(1/4)])

(-3, -1+1)

focus = (-3,0)

The last bit is the directrix. There is a formula for it, but you don't necessarily need it for this particular problem. The directrix is a line that is above or below the vertex, outside of the parabola. The vertex is always the midpoint between the focus and directrix. So we can use the midpoint formula (which I find easier to remember) instead.

so vertex=[(focus x value+directrix x value)/2, (focus y value+directrix y value)/2]

1. Plug in what you know: (-3,-1) = [(-3+x)/2,(0+y)/2]
2. make two equations, one for the x, and one for the y.
3. -3=(-3+x)/2 and -1=(0+y)/2
4. Solve
5. x=-3, y=-2
6. This value is a point on the directrix. Since the directrix is a line, we can rewrite this as an equation.

directrix: y=-2, since we have a horizontal line.

Hope this helps! Let me know if you have any questions.