How to find the equation of a quadratic function from
its graph
By Murray Bourne, 17 May 2011
http://www.intmath.com/blog/how-to-find-the-equation-of-a-quadratic-function-from-itsgraph/6070
A reader recently asked:
I would like to know how to find the equation of a quadratic function from its graph, including
when it does not cut the x-axis. Thanks.
Modelling
This is a good question because it goes to the heart of a lot of "real" math. Often we have a set of
data points from observations in an experiment, say, but we don’t know the function that passes
through our data points. (Most "text book" math is the wrong way round – it gives you the
function first and asks you to plug values into that function.)
A quadratic function’s graph is a parabola
The graph of a quadratic function is a parabola. The parabola can either be in "legs up" or "legs
down" orientation.
We know that a quadratic equation will be in the form:
y = ax2 + bx + c
Our job is to find the values of a, b and c after first observing the graph. Sometimes it is easy to
spot the points where the curve passes through, but often we need to estimate the points.
Let’s start with the simplest case. (We’ll assume the axis of the given parabola is vertical.)
Parabola cuts the graph in 2 places
We can see on the graph that the roots of the quadratic are:
x = 2 (since the graph cuts the x-axis at x = − 2); and
x = 1 (since the graph cuts the x-axis at x = 1.)
Now, we can write our function for the quadratic as follows (since if we solve the following for 0,
we’ll get our 2 intersection points):
f(x) = (x + 2)(x − 1)
We can expand this to give:
f(x) = x2 + x − 2
This is a quadratic function which passes through the x-axis at the required points.
But is this the correct answer?
It turns out there are an infinite number of parabolas passing through the points (−2,0) and (1,0).
Here are some of them (in green):
And don’t forget the parabolas in the "legs down" orientation:
So how do we find the correct quadratic function for our original question (the one in blue)?
ORR Interjection
Use y = a(x-1)(x+2)
Substitute in another point from the graph into the above equation: e.g. (0, -3)
-3 = a(0-1)(0+2)
= a (-1)(2)
= -2a
→a=
−3
−2
= 1.5
Vertex method
We can write a parabola in "vertex form" as follows:
y = a(x − h)2 + k
For this parabola, the vertex is at (h, k).
In our example above, we can’t really tell where the vertex is. It’s near(−0.5, −3.4), but "near" will
not give us a correct answer. (If there are no other "nice" points where we can see the graph
passing through, then we would have to use our estimate.)
The next example shows how we can use the Vertex Method to find our quadratic function.
One point touching the x-axis
This parabola touches the x-axis at (1, 0) only.
If we use y = a(x − h)2 + k, we can see from the graph that h = 1 and k = 0.
This gives us y = a(x − 1)2. What is the value of "a"?
But as in the previous case, we have an infinite number of parabolas passing through (1, 0). Here
are some of them:
In this example, the blue curve passes through (0, 1) on the y-axis, so we can simply
substitute x = 0, y = 1 into y = a(x − 1)2 as follows:
1 = a(− 1)2
This gives us a = 1.
So our quadratic function for this example is
f(x) = (x − 1)2 = x2 − 2x + 1
Note: We could also make use of the fact that the x-value of the vertex of the parabola y =
ax2 + bx + c is given by:
No points touching the x-axis
Here’s an example where there is no x-intercept.
We can see the vertex is at (-2, 1) and the y-intercept is at (0, 2).
We just substitute as before into the vertex form of our quadratic function.
We have (h, k) = (-2, 1) and at x = 0, y = 2.
So
y = a(x − h)2 + k
becomes
2 = a(0 − (−2))2 + 1
2 = 4a +1
a = 0.25
So our quadratic function is:
f(x) = 0.25(x −(−2))2 + 1 = 0.25(x + 2)2 + 1
f(x) = 0.25x2 + x + 2