Parabolic equations can be some of the trickiest types of equations out there. One of the most common problems students have when dealing with parabolas is when it is given in a form they don’t recognize. If you can’t understand the equation, you can’t use it properly. The two forms I will discuss are known as standard and vertex forms. Standard form is in the form of a standard quadratic equation (ie. ax^2 + bx + c) where as the vertex form is just a manipulation of the same form.
Vertex to Standard
To get an equation from vertex form to standard form you simply expand and simplify what is bracketed, known as the binomial square.
Example: Let’s use the equation
y=(x-1)^2 + 1
We start off by expanding
(x-1)^2
We are then left with
y=x^2 – 2x + 1 + 1
Finally we simplify to get our final answer in standard form:
y = x^2 – 2x + 2
Standard to Vertex
To convert to vertex form when given a equation in standard form you need to understand how to complete the square. For more experienced students, they man have no trouble with this, but for those in need of a little help there is a general formula:
(x + t/2)^2 = x^2 + 2(t/2)x + (t/2)^2
Example: (x + 3)^2 = x^2 + 2(3)x + (3)^2 = x^2 + 6x + 9
It won’t always be as simple as just completing the square. Often times you will need to add or subtract a x0 term, or a term with no variable.
Example: Let’s use the equation
y = x^2 + 8x + 13
When we first look at it, we can see that this equation does not satisfy the general formula for completing the square. In order to complete the square we must add 3to both sides.
3 + y = x^2 + 8x + 16
We then rearrange in order to isolate y on the left side.
Y = x^2 + 8x + 16 – 3
We can now complete the square using the first three terms on the right hand side.
Y = (x + 4)^2 – 3
This is now in vertex form!
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This article was written for you by Troy, one of the tutors with Test Prep Academy.