*Gerald Lemay*Show bio

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

Lesson Transcript

Instructor:
*Gerald Lemay*
Show bio

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

Quadratic equations appear in all types of science and engineering applications. In this lesson we'll look at three scenarios for writing quadratic equations when given points on the curve.
Updated: 11/09/2020

Cloud art! Ever try it? You look at clouds and you see a work of art. Then you describe what you see. Hmmm, three major shapes: related but different.

Those three different shapes are like the three forms for quadratic equations: the vertex form, the *x*-intercepts form, and the standard form. You need information to write the quadratic equation. Typically, this information is available in one of three scenarios. We are either given:

- a point and the vertex,
- a point and the
*x*-intercepts, or - three points

Three forms, three scenarios, and you the only mathematician! Let's uncloud this situation by doing examples.

Let's say we are given a point on the curve at (-1, 3) and the vertex located at (3, -1).

As you can see in this image, we use the **vertex form** of the quadratic equation:

*y* = *a*(*x* - *h*)2 + *k*.

See the vertex at (3, -1)? Since the general way to write the vertex is (*h*, *k*), we compare (*h*, *k*) to (3, -1). This means *h* = 3 and *k* = -1. Substituting these values for *h* and *k* in *y* = *a*(*x* - *h*)2 + *k* we get *y* = *a*(*x* - 3)2 - 1. Be careful of the signs! You don't want any rain on your results. It's looking good but we still need the value for *a*.

- Use the given point (-1, 3), which says
*y*is 3 for*x*equal to -1. Substituting for*x*and*y*: 3 =*a*(-1 - 3)2 - 1 => 3 =*a*(-4)2 - 1. - Now, simplify and solve for
*a*: 3 = 16*a*- 1 =>*a*= (3+1)/16 = 4/16 = 1/4 = 0.25.

Thus, our quadratic equation in vertex form is *y* = 0.25(*x* - 3)2 - 1.

What have we learned so far? When the vertex is given, we use the vertex form. One of those cloud shapes has been clarified.

In the next scenario we are given the *x*-intercepts and one point on the curve. The point is (6, 5/4) and the *x*-intercepts are located at (1, 0) and (5, 0).

The quadratic form to use is the ** x-intercepts form**, which is, as you can see in this image:

*y* = *a*(*x* - *p*)(*x* - *q*).

See the *p* and *q* in the equation? Those are the *x*-intercept values. From (1, 0) we want the *x* value which is 1. Same for (5, 0); just the *x* value. Those *x* values are the *p* and the *q*. That means *p* is 1 and the *q* is 5. Substituting the 1 and 5 in the equation gives us *y* = *a*(*x* - 1)(*x* - 5). Once again, we still need to find the *a*.

- Use the given point (6, 5/4), which says
*y*= 5/4 when*x*= 6. Substituting 6 for*x*and 5/4 for*y*in the equation: 5/4 =*a*(6 - 1)(6 - 5). - Now, simplify: 5/4 =
*a*(5)(1). - Solving for
*a*:*a*= 5/4 divided by 5 = 1/4 = 0.25.

We are done! The quadratic equation is *y* = 0.25(*x* - 1)(*x* - 5).

Two cloud shapes down and one to go. And it's still a nice day!

In the next example that you can see here, we're given three points located at (-1, 3), (4, -3/4) and (6, 5/4). No *x*-intercepts and no vertex. What do you do?

With three points, we use the standard form for the quadratic equation:

*y* = *ax*2 + *bx* + *c*.

How do we find the *a*, *b* and *c*? First, note each point tells us an *x* value and its corresponding *y* value. For example, the first point with (-1, 3) tells us *y* is 3 when *x* is -1. When we substitute -1 for *x* in the standard form equation, the *y* value will be 3. Let's do this substitution. We get 3 = *a*(-1)2 + *b*(-1) + *c*. This simplifies to 3 = *a* - *b* + *c*. Do the same with the other points and we'll get a total of three equations:

- (-1, 3) gives us:
*a*-*b*+*c*= 3. - (4, -3/4) gives us: 16
*a*+ 4*b*+*c*= -3/4. - (6, 5/4) gives us: 36
*a*+ 6*b*+*c*= 5/4.

We can take any two of these equations, subtract one from the other, and the *c* will cancel leaving us with two unknowns. This is a pretty nice feature of these equations. By subtracting two pairs of equations (your choice of which pairs), the problem reduces to finding two unknowns with two equations. We then go back and find *c*. Here are the details:

- Subtract
*a*-*b*+*c*= 3 from 16*a*+ 4*b*+*c*= -3/4.- This gives us 15
*a*+ 5*b*= -15/4, which reduces to 3*a*+*b*= -3/4.

- This gives us 15
- Subtract 16
*a*+ 4*b*+*c*= -3/4 from 36*a*+ 6*b*+*c*= 5/4.- This gives: 20
*a*+ 2*b*= 2, which reduces to 10*a*+*b*= 1.

- This gives: 20

Now we have two equations with unknowns *a* and *b*:

- 3
*a*+*b*= -3/4 and 10*a*+*b*= 1.

Subtract one from the other: 7*a* = 7/4 => *a* = 1/4.

From 3*a* + *b* = -3/4 with *a* = 1/4: 3/4 + *b* = -3/4 => *b* = -3/2.

From *a* - *b* + *c* = 3: 1/4 - (-3/2) + *c* = 3 => *c* = 5/4.

We could also solve these three equations using a matrix-vector approach. Here are the details:

Using either solution method we find *a* = 1/4 = 0.25, *b* = -3/2 = -1.5 and *c* = 5/4 = 1.25. Substituting for *a*, *b* and *c* in the standard quadratic form gives:

*y* = 0.25*x*2 - 1.5*x* + 1.25.

That last cloud shape was fun and you see clear skies ahead.

There are three typical scenarios when writing a quadratic equation from points. We are given a point and the vertex, a point and the *x*-intercepts, or three points. These scenarios relate directly to a particular form of the quadratic equation:

*y*=*a*(*x*-*h*)2 +*k*(or the**vertex form**where (*h*,*k*) is the vertex location).*y*=*a*(*x*-*p*)(*x*-*q*) (or thewhere*x*-intercepts form*p*and*q*are the*x*-intercepts).*y*=*ax*2 +*bx*+*c*(or the**standard form**).

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