How to find standard form of a parabola

(finding the equation of a parabola)

We learn how to find the equation of a parabola by writing it in vertex form

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(finding the equation of a parabola)
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Video transcript
How do you find standard form?
How do you find the standard form of a parabola given two points?

In the previous section, we learnt how to write a parabola in its vertex form and saw that a parabola's equation: \[y = ax^2+bx+c\] could be re-written in vertex form: \[y = a\begin{pmatrix}x - h \end{pmatrix}^2+k\] where:

\(h\): is the horizontal coordinate of the vertex
\(k\): is the vertical coordinate of the vertex.

This is illustrated here:

We can use the vertex form to find a parabola's equation. The idea is to use the coordinates of its vertex (maximum point, or minimum point) to write its equation in the form \(y=a\begin{pmatrix}x-h\end{pmatrix}^2+k\) (assuming we can read the coordinates \(\begin{pmatrix}h,k\end{pmatrix}\) from the graph) and then to find the value of the coefficient \(a\).

This is explained in the step-by-stem mathod, below, as well as in the tutorials.

See the explanation section.

The term "standard form" is perhaps overused in mathematics.

The standard form for a quadratics function (as a polynomial function) is #f(x)=ax^2+bx+c#.

The standard for for the equation of a parabola (also called the vertex form) is like the standard form for other conic sections.

For a parabola with vertex #(h,k)# through the points #(h+-1,k+a)#,
(that is, it opens up or down)
the standard form is #y = a(x-h)^2 + k#

(If the parabola opens sideways, it includes the points #(h+a, k+-1)# and has form #x = a(y-k)^2+h#.)

When the equation of a parabola appears in standard form, you have all the information you need to graph it or to determine some of its characteristics, such as direction or size.

Not all equations come packaged that way, though. You may have to do some work on the equation first to be able to identify anything about the parabola.

The standard form of a parabola is (x – h)2 = a(y– k) or (y – k)2 = a(x–h), where (h, k) is the vertex.

The methods used here to rewrite the equation of a parabola into its standard form also apply when rewriting equations of circles, ellipses, and hyperbolas. The standard forms for conic sections are factored forms that allow you to immediately identify needed information. Different algebra situations call for different standard forms — the form just depends on what you need from the equation.

For instance, if you want to convert the equation x2 + 10x – 2y + 23 = 0 into the standard form, you perform the following steps, which contain a method called completing the square (a method you use to solve quadratic equations):

Rewrite the equation with the x2 and x terms (or the y2 and y terms) on one side of the equation and the rest of the terms on the other side.

x2 + 10x = 2y – 23
Add a number to each side to make the side with the squared term into a perfect square trinomial (thus completing the square).
In this case, you add 25 to each side. x2 + 10x + 25 = 2y – 23 + 25 simplifies to x2 + 10x + 25 = 2y + 2.
Rewrite the perfect square trinomial in factored form, and factor the terms on the other side by the coefficient of the variable.
(x + 5)2 = 2(y + 1)

You now have the equation in standard form. The vertex is at (–5, –1); if you were to graph it, you would see that it opens upward and is fairly wide.

About This Article

This article is from the book:

Algebra II For Dummies ,

About the book author:

Mary Jane Sterling is the author of Algebra I For Dummies, Algebra Workbook For Dummies, and many other For Dummies books. She taught at Bradley University in Peoria, Illinois for more than 30 years, teaching algebra, business calculus, geometry, and finite mathematics.

This article can be found in the category:

Algebra ,

Video transcript

I have an equation right here. It's a second degree equation. It's a quadratic. And I know its graph is going to be a parabola. Just as a review, that means it looks something like this or it looks something like that. Because the coefficient on the x squared term here is positive, I know it's going to be an upward opening parabola. And I am curious about the vertex of this parabola. And if I have an upward opening parabola, the vertex is going to be the minimum point. If I had a downward opening parabola, then the vertex would be the maximum point. So I'm really trying to find the x value. I don't know actually where this does intersect the x-axis or if it does it all. But I want to find the x value where this function takes on a minimum value. Now, there's many ways to find a vertex. Probably the easiest, there's a formula for it. And we talk about where that comes from in multiple videos, where the vertex of a parabola or the x-coordinate of the vertex of the parabola. So the x-coordinate of the vertex is just equal to negative b over 2a. And the negative b, you're just talking about the coefficient, or b is the coefficient on the first degree term, is on the coefficient on the x term. And a is the coefficient on the x squared term. So this is going to be equal to b is negative 20. So it's negative 20 over 2 times 5. Well, this is going to be equal to positive 20 over 10, which is equal to 2. And so to find the y value of the vertex, we just substitute back into the equation. The y value is going to be 5 times 2 squared minus 20 times 2 plus 15, which is equal to let's see. This is 5 times 4, which is 20, minus 40, which is negative 20, plus 15 is negative 5. So just like that, we're able to figure out the coordinate. This coordinate right over here is the point 2, negative 5. Now it's not so satisfying just to plug and chug a formula like this. And we'll see where this comes from when you look at the quadratic formula. This is the first term. It's the x value that's halfway in between the roots. So that's one way to think about it. But another way to do it, and this probably will be of more lasting help for you in your life, because you might forget this formula. It's really just try to re-manipulate this equation so you can spot its minimum point. And we're going to do that by completing the square. So let me rewrite that. And what I'll do is out of these first two terms, I'll factor out a 5, because I want to complete a square here and I'm going to leave this 15 out to the right, because I'm going to have to manipulate that as well. So it is 5 times x squared minus 4x. And then I have this 15 out here. And I want to write this as a perfect square. And we just have to remind ourselves that if I have x plus a squared, that's going to be x squared plus 2ax plus a squared. So if I want to turn something that looks like this, 2ax, into a perfect square, I just have to take half of this coefficient and square it and add it right over here in order to make it look like that. So I'm going to do that right over here. So if I take half of negative 4, that's negative 2. If I square it, that is going to be positive 4. I have to be very careful here. I can't just willy nilly add a positive 4 here. I have equality here. If they were equal before adding the 4, then they're not going to be equal after adding the 4. So I have to do proper accounting here. I either have to add 4 to both sides or I should be careful. I have to add the same amount to both sides or subtract the same amount again. Now, the reason why I was careful there is I didn't just add 4 to the right hand side of the equation. Remember, the 4 is getting multiplied by 5. I have added 20 to the right hand side of the equation. So if I want to make this balance out, if I want the equality to still be true, I either have to now add 20 to y or I have to subtract 20 from the right hand side. So I'll do that. I'll subtract 20 from the right hand side. So I added 5 times 4. If you were to distribute this, you'll see that. I could have literally, up here, said hey, I'm adding 20 and I'm subtracting 20. This is the exact same thing that I did over here. If you distribute the 5, it becomes 5x squared minus 20x plus 20 plus 15 minus 20. Exactly what's up here. The whole point of this is that now I can write this in an interesting way. I could write this as y is equal to 5 times x minus 2 squared, and then 15 minus 20 is minus 5. So the whole point of this is now to be able to inspect this. When does this equation hit a minimum value? Well, we know that this term right over here is always going to be non-negative. Or we could say it's always going to be greater than or equal to 0. This whole thing is going to hit a minimum value when this term is equal to 0 or when x equals 2. When x equals 2, we're going to hit a minimum value. And when x equals 2, what happens? Well, this whole term is 0 and y is equal to negative 5. The vertex is 2, negative 5.

How do you find standard form?

The standard form of a linear equation is Ax+By=C. To change an equation written in slope-intercept form (y=mx+b) to standard form, you must get the x and y on the same side of the equal sign and the constant on the other side.