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Quadratic Functions and Factoring
Graphs of Quadratic Functions
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Algebra
Concept Version 9
Created by Boundless

Graphing Quadratic Equations in Vertex Form

The vertex form of a quadratic function lets its vertex be found easily.

Learning Objective

  • Explain the meanings of the constants aaa, hhh, and kkk for a quadratic equation in vertex form


Key Points

    • An important form of a quadratic function is vertex form: f(x)=a(x−h)2+kf(x) = a(x-h)^2 + kf(x)=a(x−h)​2​​+k
    • When written in vertex form, it is easy to see the vertex of the parabola at (h,k)(h, k)(h,k).
    • It is easy to convert from vertex form to standard form.
    • It is more difficult, but still possible, to convert from standard form to vertex form. The process involves a technique called completing the square. 

Terms

  • constant

    An identifier that is bound to an invariant value.

  • vertex

    A point on the curve with a local minimum or maximum of curvature.

  • quadratic

    A polynomial of degree two.


Full Text

Quadratic equations may take various forms. You have already seen the standard form: 

f(x)=ax2+bx+cf(x)=a{ x }^{ 2 }+bx+cf(x)=ax​2​​+bx+c

Another common form is called vertex form, because when a quadratic is written in this form, it is very easy to tell where its vertex is located. The vertex form is given by:

 f(x)=a(x−h)2+kf(x)=a(x-h)^2+kf(x)=a(x−h)​2​​+k

The vertex is (h,k).(h,k).(h,k). Note that if the form were f(x)=a(x+h)2+kf(x)=a(x+h)^2+kf(x)=a(x+h)​2​​+k, the vertex would be (−h,k).(-h,k).(−h,k). The coefficient aaa as before controls whether the parabola opens upward or downward, as well as the speed of increase or decrease of the parabola. 

Converting From Vertex Form to Standard Form

If you want to convert a quadratic in vertex form to one in standard form, simply multiply out the square and combine like terms. For example, the quadratic

y=(x−2)2+1y=(x-2)^2+1y=(x−2)​2​​+1 

Can be rewritten as follows:

y=(x−2)(x−2)+1=x2−2x−2x+4+1=x2−4x+5\begin{aligned} y&=(x-2)(x-2)+1 \\ &=x^2-2x-2x+4+1 \\ &=x^2-4x+5 \end{aligned}​y​​​​​=(x−2)(x−2)+1​=x​2​​−2x−2x+4+1​=x​2​​−4x+5​​

Converting From Standard Form to Vertex Form 

It is more difficult to convert from standard form to vertex form. The process is called "completing the square."  

Conversion When a=1a=1a=1

Consider the following example: suppose you want to write y=x2+4x+6y=x^2+4x+6y=x​2​​+4x+6 in vertex form. Note that the coefficient on x2x^2x​2​​ (the one we call aaa) is 111. When this is the case, we look at the coefficient on xxx (the one we call bbb) and take half of it. Then we square that number. Thus for this example, we divide 444 by 222 to obtain 222 and then square it to obtain 444. 

We then both add and subtract this number as follows: 

y=(x2+4x+4)+6−4y=(x^2+4x+4)+6-4y=(x​2​​+4x+4)+6−4

Note that we both added and subtracted 4, so we didn't actually change our function. Now the expression in the parentheses is a square; we can write y=(x+2)2+2.y=(x+2)^2+2.y=(x+2)​2​​+2. Our equation is now in vertex form and we can see that the vertex is (−2,2).(-2,2).(−2,2). 

Conversion When a≠1a \neq 1a≠1

It is slightly more complicated to convert standard form to vertex form when the coefficient aaa is not equal to 111. We can still use the technique, but must be careful to first factor out the aaa as in the following example: 

Consider y=2x2+12x+5.y=2x^2+12x+5. y=2x​2​​+12x+5. We factor out the coefficient 222 from the first two terms, writing this as:

y=2(x2+6x)+5y=2(x^2+6x) + 5y=2(x​2​​+6x)+5 

We then complete the square within the parentheses. Note that half of 666 is 333 and 32=93^2=93​2​​=9. So we add and subtract 999 within the parentheses, obtaining:

y=2(x2+6x+9−9)+5y=2(x^2+6x+9-9)+5y=2(x​2​​+6x+9−9)+5 

We can then finish the calculation as follows:

y=2((x+3)2−9)+5=2(x+3)2−18+5=(x+3)2−13\begin{aligned} y&=2((x+3)^2-9)+5 \\ &=2(x+3)^2-18+5 \\ &=(x+3)^2-13 \end{aligned}​y​​​​​=2((x+3)​2​​−9)+5​=2(x+3)​2​​−18+5​=(x+3)​2​​−13​​ 

So the vertex of this parabola is (−3,−13).(-3,-13).(−3,−13).

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