how to find x intercepts of a quadratic function

A reader recently asked:

I would like to know how to find the equation of a quadratic part from its graph, including when it does not cut the x-axis. Thank you.

Modeling

This is a skillful question because it goes to the middle of a lot of "real" math. Oftentimes nosotros accept a set of information points from observations in an experiment, say, only we don't know the part that passes through our information points. (Most "text book" math is the wrong way circular - information technology gives you lot the function first and asks you to plug values into that function.)

A quadratic function'south graph is a parabola

The graph of a quadratic function is a parabola. The parabola can either be in "legs upwardly" or "legs down" orientation.

Nosotros know that a quadratic equation will be in the form:

y = ax ² + bx + c

Our job is to find the values of a , b and c after first observing the graph. Sometimes information technology is easy to spot the points where the curve passes through, but often we need to judge the points.

Permit'due south kickoff with the simplest case. (We'll presume the axis of the given parabola is vertical.)

Parabola cuts the graph in 2 places

We tin see on the graph that the roots of the quadratic are:

x = −2 (since the graph cuts the x-axis at 10 = − ii); and

ten = 1 (since the graph cuts the x -axis at x = 1.)

At present, we tin can write our function for the quadratic as follows (since if we solve the following for 0, we'll get our ii intersection points):

f(10) = (x + 2)(x − ane)

Nosotros can expand this to give:

f(x) = ten ² + x − 2

This is a quadratic role which passes through the x-centrality at the required points.

Just is this the correct respond?

Observe my graph passes through −three on the y-centrality. Allow'due south substitute 10 = 0 into the equation I but got to check if it'south right.

f(0) = 0² + 0 − 2 = −ii

It's non correct!

It turns out there are an space number of parabolas passing through the points (−2,0) and (1,0).

Here are some of them (in light-green):

And don't forget the parabolas in the "legs down" orientation:

So how practice we observe the right quadratic function for our original question (the ane in blueish)?

System of Equations method

To notice the unique quadratic function for our blue parabola, we need to use 3 points on the bend. We can so class iii equations in 3 unknowns and solve them to get the required result.

On the original blue curve, we can run across that information technology passes through the point (0, −3) on the y-axis. We'll utilise that as our 3rd known signal.

Using our full general course of the quadratic, y = ax ⁱⁱ + bx + c , we substitute the known values for x and y to obtain:

Substituting (−2,0):

0 = a(−2)² + b(−ii) + c = 4a − ii b + c

Substituting (one,0):

0 = a(1)² + b(one) + c = a + b + c

Substituting (0,−3):

−three = a(0)² + b(0) + c = c

So nosotros get c = −3.

Substituting c = −3 in the offset line gives:

foura − 2b = three; and substituting into the second line gives:

a + b = 3

Multiplying the concluding line by 2 gives:

2a + 2b = 6

Adding this to iva − 2b = 3 gives:

6a = 9

This gives a = i.v.

Substituting a = 1.5 into a + b = 3, nosotros get b = i.5.

So the correct quadratic function for the bluish graph is

f(x) = i.vx ² + i.5ten − 3

We note that the " a " value is positive, resulting in a "legs up" orientation, every bit expected.

Vertex method

Some other way of going about this is to observe the vertex (the "pointy end") of the parabola.

Nosotros can write a parabola in "vertex course" as follows:

y = a(x − h)² + chiliad

For this parabola, the vertex is at (h, yard).

In our case above, nosotros can't actually tell where the vertex is. Information technology'southward almost (−0.5, −three.iv), but "almost" volition non requite us a correct answer. (If there are no other "squeamish" points where we can see the graph passing through, then nosotros would have to utilize our approximate.)

The next example shows how we can use the Vertex Method to discover our quadratic function.

One point touching the ten-centrality

This parabola touches the 10-axis at (one, 0) merely.

If we use y = a(x − h)² + k , we can see from the graph that h = i and thousand = 0.

This gives u.s.a. y = a(x − 1)² . What is the value of "a"?

Merely as in the previous instance, we have an space number of parabolas passing through (ane, 0). Here are some of them:

In this example, the bluish curve passes through (0, 1) on the y-axis, so nosotros tin can merely substitute x = 0, y = 1 into y = a(x − i)² as follows:

ane = a(− one)²

This gives u.s.a. a = 1.

Then our quadratic function for this case is

f(10) = (x − 1)² = 10 ⁱⁱ − 2ten + one

Notation: We could also make utilize of the fact that the x-value of the vertex of the parabola y = ax ² + bx + c is given past:

No points touching the x-axis

Here's an example where there is no x-intercept.

Nosotros can see the vertex is at (-2, one) and the y-intercept is at (0, ii).

We just substitute as before into the vertex class of our quadratic function.

We have (h, k) = (-2, i) and at x = 0, y = 2.

So

y = a(10 − h)² + k

becomes

two = a(0 − (−2))² + 1

ii = 4a +ane

a = 0.25

And then our quadratic function is:

f(x) = 0.25(x −(−2))^two + ane = 0.25(x + ii)² + 1 = 0.25(ten ² + 410 + iv) + 1

f(x) = 0.25x ² + ten + 2

Using math software to find the function

a. Wolfram|Alpha

This Wolfram|Alpha search gives the answer to my terminal example.

b. Excel

You could utilize MS Excel to find the equation. Enter the points in cells equally shown, and get Excel to graph it using "X-Y scatter plot". This gives the black curve shown. And so right click on the curve and cull "Add trendline" Choose "Polynomial" and "Society two". (This gives the blue parabola as shown below).

In the "Options" tab, choose "Display equation on chart".

We go the post-obit result.

c. GeoGebra

GeoGebra was not then useful for this task. GeoGebra will give us the equation of a parabola, but yous need to know the focus and directrix first. This is not so straightforward from observations of a graph.

Conclusion

Finding the equation of a parabola given certain data points is a worthwhile skill in mathematics. Parabolas are very useful for mathematical modelling considering of their simplicity.

Encounter the 82 Comments below.

Source: https://www.intmath.com/blog/mathematics/how-to-find-the-equation-of-a-quadratic-function-from-its-graph-6070

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