1. Graphing Quadratic Functions Using Vertex Form and Intercept Form
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Chapter 8
1.
Graphing Quadratic Functions Using Vertex Form and Intercept Form
Quadratic functions can be written in two different forms: vertex form and intercept form. Vertex form shows the vertex of the parabola, while intercept form shows the x- and y-intercepts of the parabola.
To convert a quadratic function from standard form to vertex form, complete the square. To convert a quadratic function from standard form to intercept form, factor out the x and y terms.
To graph a quadratic function in vertex form, start by finding the vertex. Then, find the x- and y-intercepts. Once you have found the vertex and the intercepts, sketch the parabola.
Vertex form is often easier to use when finding the vertex of the parabola, while intercept form is often easier to use when finding the x- and y-intercepts of the parabola.
The best way to learn how to graph quadratic functions is to practice. There are many online resources that can help you practice graphing quadratic functions.
Here are some tips for graphing quadratic functions:
Start by finding the vertex of the parabola.
Then, find the x- and y-intercepts of the parabola.
Once you have found the vertex and the intercepts, sketch the parabola.
The parabola will be symmetric about the vertical line that passes through the vertex.
The parabola will open up if the coefficient of x² is positive, and it will open down if the coefficient of x² is negative.
Show less Show more expand_more Lesson Settings & Tools
| | 17 Theory slides |
| | 8 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
This lesson will discuss how to graph a particular type of function called a quadratic function.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Explore
Maximum and Minimum Points of Graphs
In the coordinate plane, the graphs of three functions and their corresponding equations are shown.
Pay close attention to the maximum/minimum point of each graph. Can their coordinates be identified just by looking at the corresponding equation?
Discussion
Parabola and Quadratic Function
All three graphs of the functions presented earlier have a very distinctive form. These curves have a specific name, which now will be properly introduced.
Concept
Parabola
A parabola is a curve that is geometrically defined as the locus of all points equidistant from a line and a point not on the line. The line is called the directrix, and the point is the focus of the parabola. In the following applet, point P is equidistant from the directrix d and the focus F. 
A parabola can be vertical or horizontal. A vertical parabola can open upward or downward. In comparison, a horizontal parabola can open to the left or the right.
Equations of parabolas always contain a variable raised to the second power. This is why the functions that represent vertical parabolas are called quadratic functions.
Concept
Quadratic Function
A quadratic function is a polynomial function of degree 2 that can always be written in the form y=ax2+bx+c, with a=0. Notice that the highest exponent of the independent variable is 2. The graph of any quadratic function is a vertical parabola.
There are three ways to write a quadratic function.
| Name | Equation | Characteristics |
|---|---|---|
| Standard Form | y=ax2+bx+c | a, b, and c are real numbers, a=0, and c is the y-intercept of the parabola. |
| Vertex Form | y=a(x−h)2+k | a, h, and k are real numbers, a=0, and (h,k) is the vertex of the parabola. |
| Intercept Form (also called Factored Form) |
y=a(x−x1)(x−x2) | a, x1, and x2 are real numbers, a=0, and x1 and x2 are the x-intercepts of the parabola. |
Discussion
Characteristics of Quadratic Functions
The inherent shape of parabolas gives rise to several characteristics that all quadratic functions have in common.
Concept
Direction of a Parabola
A parabola either opens upward or downward. This is the direction of the parabola . If the leading coefficient a of the corresponding equation is positive, the parabola opens upward. If the coefficient is negative, the parabola opens downward.
Concept
Vertex of a Parabola
Because a parabola either opens upward or downward, there is always one point that is the absolute maximum or absolute minimum of the function. This point is called the vertex.
At the vertex, the function changes from increasing to decreasing, or vice versa.
Concept
Axis of Symmetry
An axis of symmetry is the line that divides the graph of a function in two mirrored images. For example, the graph of a quadratic function is a parabola that has an axis of symmetry parallel to the y-axis and passes through its vertex.
It is also possible for a function to have an axis of symmetry which is not vertical. The following graph shows an example of this. 
If a function's axis of symmetry is the y-axis, it is said that it is an even function.
The concept of an axis of symmetry extends beyond graphs of functions.
Any geometric figure can have an axis of symmetry if there exists a line that divides it into congruent, mirror-image halves.Concept
Zeros of a Parabola
A parabola can intersect the x-axis at zero, one, or two points. Since the function's value at an x-intercept is always 0, these points are called zeros, or sometimes roots.
y-intercept
Because all graphs of quadratic functions extend infinitely to the left and right, they each have a y-intercept somewhere along the y-axis.
Pop Quiz
Identifying Characteristics of a Parabola
Consider the given parabola. Identify its zeros, line of symmetry, or y-intercept.
Discussion
Vertex Form of a Quadratic Function
A quadratic function is said to be written in vertex form if it has the following format.
y=a(x−h)2+k
Here, a, h, and k are real numbers with a=0. The value of a gives the direction of the parabola. When a>0, the parabola faces upward, and when a<0, it faces downward. The vertex of the parabola lies at (h,k), and the axis of symmetry is the vertical line x=h. Consider the graph of y=-41(x−4)2+8.
Vertex Form:Example Function:y= a(x−h)2+ky=-41(x−4)2+8
These values determine the characteristics of the parabola shown in the graph. | Direction | Vertex | Axis of Symmetry |
|---|---|---|
| a=-41 | h=4 and k=8 | h=4 |
| Since -41 is less than 0, the parabola opens downward. | The vertex is located at (4,8). | The axis of symmetry is the vertical line x=4. |
Extra
Consider other example quadratic functions.
Function 1:Function 2:Function 3:y=2(x+1)2+7y=2(x−3)2+1y=5(x−2)2−3
Although these functions do not strictly follow the format for the vertex form, they are said to be written in vertex form because they can easily be rewritten in the desired format. | Function 1 | Function 2 | Function 3 |
|---|---|---|
| y=2(x+1)2+7 ⇕ y=2(x−(-1))2+7 |
y=(x−3)2+1 ⇕ y=1(x−3)2+1 |
y=5(x−2)2−3 ⇕ y=5(x−2)2+(-3) |
Discussion
Graphing a Quadratic Function in Vertex Form
When a quadratic function is written in vertex form, some characteristics of its graph can be identified.
| y=a(x−h)2+k | ||
|---|---|---|
| Direction | Vertex | Axis of Symmetry |
| a>0 a<0 ⇒ upward⇒ downward
|
(h,k) | x=h |
1
Identify and Plot the Vertex
expand_more To begin, identify the vertex (h,k) from the function rule.
y=(x−2)2−4⇕y=(x−2)2+(-4)
Here, h=2 and k=-4. Therefore, the vertex of the parabola is (2,-4). This point can be plotted on a coordinate plane. 2
Draw the Axis of Symmetry
expand_more The axis of symmetry is a vertical line that passes through the vertex and divides the parabola into two mirror images. Since the axis of symmetry is x=h, here it is x=2.
3
Determine and Plot the y-intercept
expand_more Next, determine the y-intercept. Recall that for all y-intercepts, the x-coordinate is 0. Therefore, substitute x=0 into the function rule and solve for y.
The y-intercept of the parabola is (0,0). This point can be added to the graph. 
4
Reflect the y-intercept Across the Axis of Symmetry
expand_more The axis of symmetry divides the parabola into two mirror images. Therefore, points on one side of the parabola can be reflected across the axis of symmetry.
5
Draw the Parabola
expand_more Now, with three points plotted, the direction of the parabola can be seen. It appears that the parabola faces upward. Identify the value of a in the equation to see if this is correct.
y=(x−2)2−4⇕y=1(x−2)2−4
Since a=1, which is greater than zero, it can be said that the parabola opens upward. To graph the quadratic function, connect the three points with a smooth curve. Example
Using the Vertex Form to Draw a Parabola
LaShay loves playing golf.
y=-2(x−23)2+29
Draw the parabola to help LaShay improve her swing!
Answer
Solution
To identify the vertex of the parabola, the given equation can be compared to the general form of a quadratic function written in vertex form.
The y-intercept will now be determined and plotted. To do so, x=0 will be substituted in the given equation.
The parabola intercepts the y-axis at (0,0). 
Next, the y-intercept will be reflected across the axis of symmetry.
Note that for the given function, the value of a is -2. Therefore, the parabola opens downward. This corresponds to the loci of the points plotted in the coordinate plane. Finally, these points can be connected with a smooth curve to draw the parabola. 
General Formy=a(x−h)2+kGiven Functiony=-2(x−23)2+29
For the given function, h=23 and k=29. This means that the vertex of the parabola is (23,29). Since the parabola illustrates the position of the ball, the first quadrant will only be considered. The axis of symmetry of the parabola is the vertical line through the vertex. Therefore, in this case, the equation of the axis of symmetry is x=23.
y=-2(x−23)2+29
Substitute
x=0
y=-2(0−23)2+29
▼
Evaluate right-hand side
SubTerm
Subtract term
y=-2(-23)2+29
NegBaseToPosPow
(-a)2=a2
y=-2(23)2+29
PowQuot
(ba)m=bmam
y=-2(49)+29
MoveLeftFacToNum
a⋅cb=ca⋅b
y=-418+29
ReduceFrac
ba=b/2a/2
y=-29+29
AddFrac
Add fractions
y=0
Example
Writing the Vertex Form of a Quadratic Function Given its Graph
Mark is studying parabolas so that he can help LaShay with her golf swing. He wants to write the vertex form of the quadratic function that corresponds to the given graph.
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Hint
Start by identifying the vertex of the parabola.
Solution
Recall the format of the vertex form.
The vertex is the point with coordinates (3,-2). This means that in the equation that corresponds to the given parabola, h=3 and k=-2. 
The parabola intercepts the y-axis at (0,1). Therefore, to find the value of a, x=0 and y=1 will be substituted into the equation and the equation will be solved for a.
It has been found that the value of a is 31. With this information, the equation of the given parabola can be written in vertex form.
y=a(x−h)2+k
In this format, the vertex of the parabola has coordinates (h,k). Therefore, to state the values of h and k, start by identifying the vertex. y=a(x−3)2+(-2)⇕y=a(x−3)2−2
Finally, to find the value of a, any point on the parabola can be used. For simplicity, the y-intercept will be used. y=a(x−3)2−2
SubstituteII
x=0, y=1
1=a(0−3)2−2
▼
Solve for a
SubTerm
Subtract term
1=a(-3)2−2
NegBaseToPosPow
(-a)2=a2
1=a(9)−2
AddEqn
LHS+2=RHS+2
3=a(9)
DivEqn
LHS/9=RHS/9
93=a
ReduceFrac
ba=b/3a/3
31=a
RearrangeEqn
Rearrange equation
a=31
y=31(x−3)2+(-2)⇕y=31(x−3)2−2
Explore
Zeros of a Parabola
In the coordinate plane below, the graphs of three quadratic functions and their corresponding equations are shown.
Pay close attention to the zeros of each graph. Can their coordinates be identified just by looking at the corresponding equation?
Discussion
Factored Form of a Quadratic Function
A quadratic function is said to be written in factored form, or intercept form, if it follows a specific format.
y=a(x−p)(x−q)
Here, a, p, and q are real numbers with a=0. The value of a gives the direction of the parabola. When a>0, the parabola faces upward, and when a<0, it faces downward. The zeros of the parabola are p and q, and the axis of symmetry is the vertical line x=2p+q.
Example
Consider the graph of y=21(x−7)(x−13).
Factored Form:Example Function:y= a(x−p)(x−q)y=21(x−7)(x−13)
These values determine the characteristics of the parabola shown in the graph. | Direction | Zeros | Axis of Symmetry |
|---|---|---|
| a=21 | p=7 and q=13 | 2p+q ⇓ 27+13=10 |
| Since 21 is greater than 0, the parabola opens upward. | The zeros are 7 and 13. Therefore, the parabola intersects the x-axis at (7,0) and (13,0). | The axis of symmetry is the vertical line x=10. |
Extra
Consider other example quadratic functions.
Function 1:Function 2:Function 3:y=2(x+1)(x−3)y=(x−5)(x−9)y=5x(x−2)
Although these functions do not strictly follow the format of the factored form, they are said to be written in factored form because they can easily be rewritten in the desired format. | Function 1 | Function 2 | Function 3 |
|---|---|---|
| y=2(x+1)(x−3) ⇕ y=2(x−(-1))(x−3) |
y=(x−5)(x−9) ⇕ y=1(x−5)(x−9) |
y=5x(x−2) ⇕ y=5(x−0)(x−2) |
Pop Quiz
Identifying Vertex and Factored Forms
In the following applet, several quadratic functions are expressed in different forms. Are the quadratic functions written in vertex form, factored form, or neither?
Discussion
Graphing a Quadratic Function in Factored Form
When a quadratic function is written in factored form, some characteristics of its graph can be identified.
| y=a(x−p)(x−q) | ||
|---|---|---|
| Direction | Zeros | Axis of Symmetry |
| a>0 a<0 ⇒ upward⇒ downward
|
p and q | x=2p+q |
It is possible to graph a quadratic function using these characteristics. Consider the function y=-(x+1)(x−5).
1
Identify and Plot the Zeros
expand_more To begin, identify the zeros p and q from the function rule.
y=-(x+1)(x−5)⇕y=-(x−(-1))(x−5)
Here, p=-1 and q=5. Therefore, the x-intercepts of the parabola are (-1,0) and (5,0). These points can be plotted on a coordinate plane. 2
Determine and Draw the Axis of Symmetry
expand_more The axis of symmetry is a vertical line with equation x=2p+q. By substituting the values of p and q, this equation can be evaluated.
The axis of symmetry x=2 can now be drawn. 
3
Determine and Plot the Vertex
expand_more The axis of symmetry is a vertical line that goes through the vertex of the parabola. Since the equation of the axis of symmetry is x=2, the x-coordinate of the vertex is also 2. To find its y-coordinate, x=2 will be substituted into the given equation.
The vertex lies at (2,9). This point can be added to the graph. 
y=-(x+1)(x−5)
Substitute
x=2
y=-(2+1)(2−5)
▼
Evaluate right-hand side
AddSubTerms
Add and subtract terms
y=-(3)(-3)
RemovePar
Remove parentheses
y=-3(-3)
MultNegNegOnePar
-a(-b)=a⋅b
y=9
4
Draw the Parabola
expand_more Now, with the three points plotted, the direction of the parabola can be seen. It appears that the parabola faces downward. Identify the value of a in the equation to see if this is correct.
y=-(x+1)(x−5)⇕y=-1(x+1)(x−5)
Since a=-1, which is less than zero, it can be said that the parabola opens downward. To graph the quadratic function, connect the three points with a smooth curve. Example
Using the Factored Form to Draw a Parabola
By now, it is not a secret that LaShay loves playing golf.
y=-2x(x−4)
Draw the parabola to help LaShay improve her swing!
Answer
Solution
To identify the zeros of the quadratic function, the given equation can be rewritten and compared to the general form of a quadratic function written in factored form.
Next, the parabola's axis of symmetry can be determined.
The axis of symmetry is the vertical line with equation x=2. 
Since the axis of symmetry is the vertical line through the parabola's vertex, the x-coordinate of the vertex is 2. To find its y-coordinate, x=2 can be substituted in the given equation.
The vertex of the parabola lies at (2,8). 
General Formy=a(x−p)(x−q)Given Functiony=-2x(x−4)⇕y=-2(x−0)(x−4)
For the given function, p=0 and q=4. This means that the zeros of the function are 0 and 4, so the parabola intersects the x-axis at (0,0) and (4,0). Note that, for the given function, the value of a is -2. Therefore, the parabola opens downward. This corresponds to the loci of the points plotted in the coordinate plane. Finally, these points can be connected with a smooth curve to draw the parabola.
Example
Writing the Factored Form of a Quadratic Function Given its Graph
Continuing his studies, Mark wants to write the factored form of the quadratic function that corresponds to the given parabola.
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Hint
Start by identifying the x-intercepts of the parabola.
Solution
Recall the format of the factored form of a parabola.
The parabola intersects the x-axis at (-5,0) and (1,0). This means that p and q are -5 and 1. It does not matter which value is attributed to which variable, so it will arbitrarily be considered that p=-5 and q=1. With this information, the following equation can be written.
The parabola's vertex lies at (-2,-3). Therefore, -2 and -3 can be substituted for x and y, respectively, in the partial equation. The equation can then be solved for a.
It has been found that the value of a is 31. With this information, the equation of the given parabola can be written in factored form.
y=a(x−p)(x−q)
In this format, the zeros of the function are p and q. Therefore, to state the values of p and q, start by identifying the x-intercepts of the graph. y=a(x−(-5))(x−1)⇕y=a(x+5)(x−1)
Finally, to find the value of a, any point on the parabola can be used. For simplicity, the vertex will be used. y=a(x+5)(x−1)
SubstituteII
x=-2, y=-3
-3=a(-2+5)(-2−1)
▼
Solve for a
AddSubTerms
Add and subtract terms
-3=a(3)(-3)
MultPosNeg
a(-b)=-a⋅b
-3=a(-9)
DivEqn
LHS/(-9)=RHS/(-9)
-9-3=a
DivNegNeg
-b-a=ba
93=a
ReduceFrac
ba=b/3a/3
31=a
RearrangeEqn
Rearrange equation
a=31
y=31(x−(-5))(x−1)⇕y=31(x+5)(x−1)
Pop Quiz
Identifying Vertex or Zeros From an Equation
For the following quadratic functions, identify the vertex or the zeros.
Closure
Standard Form of a Quadratic Function
The same quadratic function can be expressed in both vertex form and factored form. For example, consider the following equations.
As shown, when expanded both the vertex form and the factored form result in the following equation.
Vertex Form:Factored Form:y=-2(x−2)2+2y=-2(x−1)(x−3)
The equations above represent the same function. To prove this, both of them will be expanded.
-2(x−2)2+2=?-2(x−1)(x−3)
▼
Simplify
ExpandNegPerfectSquare
(a−b)2=a2−2ab+b2
-2(x2−2x(2)+22)+2=?-2(x−1)(x−3)
CommutativePropMult
Commutative Property of Multiplication
-2(x2−2(2)x+22)+2=?-2(x−1)(x−3)
Multiply
Multiply
-2(x2−4x+22)+2=?-2(x−1)(x−3)
CalcPow
Calculate power
-2(x2−4x+4)+2=?-2(x−1)(x−3)
Distr
Distribute -2
-2x2+8x−8+2=?(-2x+2)(x−3)
Distr
Distribute (x−3)
-2x2+8x−8+2=?-2x(x−3)+2(x−3)
Distr
Distribute -2x & 2
-2x2+8x−8+2=?-2x2+6x+2x−6
AddTerms
Add terms
-2x2+8x−6=-2x2+8x−6 ✓
y=-2x2+8x−6
This form is called the standard form of a quadratic function. For more information, read about standard form of a quadratic function.
f(x)=(x−3)2+4
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Let's start by recalling the vertex form of a quadratic function with a vertex at ( h, k). f(x)=a(x- h)^2+ k Note that the quadratic function is already given in its vertex form. This way we can easily identify the coordinates of its vertex. f(x)=(x- 3)^2+ 4 Therefore, its vertex is at (3,4).
The axis of symmetry of a quadratic function is the vertical line that divides the graph of the function in two mirrored images. When the quadratic function is written in vertex form f(x)=a(x- h)^2+ k, the axis of symmetry is given by the following equation.
x = h
We found that h=3 in Part A. This means that the axis of symmetry is the line x=3.
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Consider the following function.
f(x)=21(x−1)2−1
Which of the following is the graph of f(x)?
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To match the function with its graph, we should first identify the vertex and then determine whether the parabola opens upwards or downwards. To do so, notice that the equation of the parabola is already given in its vertex form. y= a(x- h)^2+ k [0.5em] ⇓ [0.5em] f(x)= 1/2(x- 1)^2+( -1) We can see that a= 12, h= 1, and k= - 1. Since the vertex of a quadratic function written in vertex form is the point ( h, k), the vertex of our function is ( 1, -1). Only graphs B and C have the correct vertex.
Let's now determine the direction of the parabola. Recall that if a>0, the parabola opens upwards. Conversely, if a<0, the parabola opens downwards. In the given function, we have a= 12, which is greater than 0. Therefore, the parabola opens upwards. As a result, the graph of the given function is graph B.
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Consider the given graph.
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The x-intercepts are the points where the graph intersects the x-axis. Let's take a look at our graph again.
We can see that the x-intercepts of the parabola are at x=-2 and x=0.
We begin by identifying the vertex of the parabola. The vertex of the parabola is the point at which it changes direction.
We can see that the vertex of the parabola is (-1,-1), where the parabola changes from decreasing to increasing. The axis of symmetry is the vertical line that goes through the vertex of the parabola. We can identify this line on the graph.
Therefore, the equation of the axis of symmetry is x=-1.
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f(x)=3(x−2)(x+6)
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Let's begin by reviewing the intercept form of a quadratic function. y=a(x-p)(x-q) Here, p and q are the x-intercepts of the function. Let's compare this to the given function. f(x)=3(x-2)(x+6) ⇕ f(x)=3(x- 2)(x-( -6)) Therefore, the x-intercepts of the given function are at x=2 and x=-6.
The axis of symmetry of a parabola is the vertical line located halfway between the x-intercepts. Since we found the x-intercepts are x=2 and x=-6, we will find the x-value that is halfway between these numbers.
x=p+q/2 ⇒ x=2+(-6)/2
Let's evaluate this expression.
Therefore, the axis of symmetry is at x=-2.
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Graphing Quadratic Functions Using Vertex Form and Intercept Form
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