Parabolas
Three Forms of the Equation
Each of the three forms of the equation for parabolas tell you something different, but all can be
used to graph an equation. For each form you will need to find the vertex and axis of symmetry,
determine which direction the graph goes, and perhaps a few points on the graph (like intercepts)
1. STANDARD FORM: y = ax2 + bx + c
Vertex: find the x-value of the vertex by using the handy dandy formula –b/2a
Once you find the x-value, plug it into the equation to find the y-value.
Axis of symmetry: This is the line that is smack in the middle of the parabola – it is the
mirror for a parabola. In standard form it is a vertical line in the form: x = –b/2a
y-intercept: This is an easy one – like always we find the y-intercept by plugging in x = 0.
In this case that makes the y-intercept (0, c)
Other points: Plug in an x-value on one side or another of the vertex into the equation. If
you know where the axis of symmetry is then you get some points for free by finding
reflections over the axis of symmetry. Plot a point, count horizontally to the axis of
symmetry, then keep on going the same number and place a point.
Ex: y = x2 – 4x - 5
a = 1, b = -4, c = 5
Vertex x = 4/2(1) = 2 plug in x = 2
Y = (2)2 -4(2) - 5 = 4 - 8 – 5 = -9
(2, -9)
axis of symmetry is x = 2 (through the vertex)
y – intercept (0, -5)
Plug in another point like x = -1
Which gives us (-1, 0)
Now we can count from the axis of symmetry
From our two points to get two additional points:
From (0,-5) to the axis you must travel 2 spaces to the
right, so keep going two more spaces to get the point (4, -5).
From (-1, 0) you must travel 3 spaces to the right, continue
on 3 more spaces and come to (5, 0).
Now sketch the graph through those points.
AK 11/08
2. VERTEX FORM: y = a(x- h)2 + k
If you were paying attention in chapter 1 when we graphed absolute value functions then this
form is not new to you.
The vertex is at (h, k)
Axis of symmetry at x = h
Plug in x = 0 for the y-intercept
Follow the y-intercept across the axis of symmetry.
You may need another point, but this can by found
By picking an x-value close to the vertex and plugging
into the function.
y = -1(x + 3)2 + 1
Vertex at (-3, 1) (since x + 3 = x – h)
Axis of symmetry at x = -3
y-intercept is y = -1(0 + 3)2 + 1 = -8
(0, -8)
Following this point 3 units left, the mirror point
must be at (-6, -8)
This info gives us a pretty good picture of the graph.
3. INTERCEPT FORM y = a(x – p)(x – q)
This form does not easily give up the vertex.
We can find x-intercepts at p and q, since they make
the y-value 0.
The tricky part is to find the average of these x-intercepts
to find the axis of symmetry, and thus the vertex by
plugging that x-value into the function.
y = ½(x – 2)(x + 4) gives us (2, 0) and (-4, 0) as intercepts. The average of 2 and -4 is -1.
The Vertex must be at (-1, ___) y = ½ (-1 - 2)(-1 + 4) = ½ (-3)(3) = -4.5 Vertex at (-1, -4.5)
This is enough info to graph.
AK 11/08