Presented by Dr. Del Ferster. We’ll spend a bit of time looking at some “test-type” problems....

Breakout Session #3Graphs of Quadratic Functions and Linear

Functions.Modeling Quadratic and Linear Functions

Presented byDr. Del Ferster

What’s in store for today? We’ll spend a bit of time looking at some

“test-type” problems. We’ll re-visit quadratic functions. This time,

we’ll emphasize the graphs of quadratics.◦ We’ll look at vertex and the intercepts.

We’ll spend a bit of time on the graphs of linear functions.◦ We’ll consider intercepts, and slope intercept form.

Then, we’ll look at some problems that model both kinds of functions (quadratic and linear).

Practice Problems Graphing Quadratic Functions and Linear Functions

Let’s do some problems!

Answers to practice problems

1. y=(x – 2)^2 -2, so B 2. Vertex is at (-3, 4), so C 3. C 4. y-int. (0,3): x-int. (4,0), so A

5. , so C

6. (0, 16), so D

33

4y x

Practice Problem Answers (continued)

7. (4, 0) or (2, 0), so C 8. Vertex at (3,-2), opens up, so C

9A.


9B. m=-2 9C. 9D. (0,4) or 4 9E. (2,0) or 2

10.

11A. (0,-5) or -5 11B. 5 or 1, or alternately (5, 0) or (1, 0)

(3, -2) is a minimum, so C 10. 11A. (0, 10) 11B. 5 or 1, or alternately (5, 0) or (1, 0) 11C. (3, -8)

22( 1) 3y x

2 4y x


11C. (3,-4) 11D.

Quadratic Functions

Graphs, Forms, and Intercepts

First, a picture of Conor!

The Graph of a Quadratic Function The graph of any quadratic function is called

a parabola. Parabolas are shaped like cups, as shown in

the graphs on the next slide. If the coefficient of x2 is positive, the

parabola opens upward; otherwise, the parabola opens downward.

The vertex (or turning point) is the minimum or maximum point.

Some Sample Graphs

Forms for Quadratic Functions We’ll consider several forms, each tends to

have some information that we can retrieve easily.

1. VERTEX FORM: In this form, we’re able to easily identify the vertex of the parabola. (h,k)

y = a(x - h)2 + k, a 0

Forms for Quadratic Functions 2. STANDARD FORM: This form works

well if you wish to apply the quadratic formula to find the zeros of the function. (ZEROS of a function are the places where the graph of the function crosses the x axis.

2y ax bx c

Forms for Quadratic Functions 3. FACTORED FORM: This form makes it

easy to find the zeros of the function. We just use the factors The zeros are at

1 2( )( )y a x z x z

1 2 and z z

More on the graphs of Quadratic Functions

We often seek to find the following information about a quadratic function.◦The y intercept◦The zeros◦The vertex◦Whether the graph opens upward or downward

◦The maximum or minimum value of the function (depending on how the graph opens)


Let’s look at these ideas one at a time.◦The y intercept

◦This is relatively easy to find, if we realize that the location where a graph crosses the y axis, it’s x value is ZERO.

◦So, to find a y intercept, just “plug 0 in for x”.


Let’s look at these ideas one at a time.◦The zeros◦Depending on the form, these can be easy or tougher to find. ZEROS are what most books now call the places where the graph crosses the x axis. You might have once called these x intercepts.

◦So, to find a zero, just “set y equal to 0, and get ready to solve for x”.


Let’s look at these ideas one at a time.◦The vertex

◦This is the “turning point” of the curve. It’s also where the graph reaches its maximum or minimum value (depending on whether it opens downward or upward.

◦In vertex form, this is (h,k).

More on Vertex If by chance, your function isn’t in vertex

form, but maybe instead is in standard form

In this case, the vertex has an x coordinate of

2y ax bx c

2

b

a


Let’s look at these ideas one at a time.◦Whether the graph opens upward or downward

◦This is easy to determine. We simply look at the coefficient on the square term. If it’s positive, the graph opens UPWARD

If it’s negative, the graph opens DOWNWARD


Let’s look at these ideas one at a time.◦ The maximum or minimum value of the

function (depending on how the graph opens)

◦ This is easy to determine, and it depends on whether the graph opens upward or downward. In either case, the max/min value of the function is the y value at the vertex. If the graph opens UPWARD we have a minimum

If the graph opens DOWNWARD we have a maximum.

Let’s do a problem together

For the function

A. Find the y intercept B. Find the zeros C. Find the vertex D. Determine whether the graph opens

upward or downward E. Find the maximum or minimum value of

the function.

22( 1) 8y x

Example: Finding the y intercept

To find the y intercept: Plug 0 in for x

22( 1) 8y x

22(0 1) 8y 22( 1) 8y

22(1) 8 2 8y 6y

So, our y intercept is

(0, -6)

Example: Finding the zeros

To find the zeros: Set y equal to zero and solve for x

22( 1) 8y x

20 2( 1) 8x 28 2( 1)x

24 ( 1)x 2 1 3 or 1x x x

So, our zeros are at

3 or -1

Example: Finding the vertex

This is easy to do, since we’re given vertex form:

The vertex is (h, k)

22( 1) 8y x

2( )y a x h k 1 (note h is positive 1)h

8k

So, our vertex is

(1, -8)

Example: Determine how the graph opens

To the opening direction, just look at the number in front.

In this case it’s a 2 (positive)

22( 1) 8y x

So, our graph opens UPWARD

Example: Determine the max/min value of the function

In this case, since the graph opens UPWARD, we have a minimum

It occurs at the vertex (which is at (1, -8)

22( 1) 8y x

So, our function has a minimum value of -8

The final graph with our information

Y intercept(0, -6)

zeros are at3 or -1

vertex is(1, -8)

graph opens UPWARD

So, our function has a minimum value of -8

Quadratic Functions

A closer look at Quadratic FunctionsVERTEX FORMSTANDARD FORM

We’ll look at the process of converting a quadratic function from standard form to vertex form.

Then, just for kicks, we’ll look at the process for converting a quadratic function from vertex form to standard form.

It has been my experience, that students usually prefer vertex form, especially those students who are good at shifts.

Overview of the section

A couple of thoughts:◦ If the graph of the parabola crosses the x axis

at ONLY one point, that point is the vertex.◦ If the graph of the parabola crosses the x axis

at 2 points, the vertex falls in the middle of these two x values. In this case, we can find the zeros, then

“average them” to determine the x value of the vertex.

◦ If there are no zeros, YIKES…finding the vertex in this case is a bit tougher.

Finding the Vertex from Standard Form

Although it’s not one of my favorite mathematical ideas, there is a “formula” that allows us to locate the vertex, if we have a quadratic function in standard form.

Recall that standard form is:◦ The vertex is located at the point

Finding vertex (given standard form)

2( )f x ax bx c

,2 2

b bf

a a

,2 2

b bf

a a

Breaking down that last formula

2( )f x ax bx c The general form:

The formula for vertex:

To find the x coordinate of the vertex:

To find the y coordinate of the vertex

2

bx

a

Plug the above x value into the function, and get a y value to go along with it

The general form:

The formula for vertex:

To find the x coordinate of the vertex:

To find the y coordinate of vertex

Find the vertex, then sketch the graph:

Example 1

2( ) 6 20f x x x

1

6

a

b

( 6)

2(1)x

63

2x

Plug the 3 into the function, we get a y value of 9-18+20 or y = 11

3,11The Vertex

x

y

The graph for that function

The Vertex 3,11

Recall that vertex form looks like this:

While standard form looks like this:

Converting from Vertex form to Standard form

2( )f x ax bx c

2( )f x a x h k

Really, it’s not hard to do at all; we just “multiply out” the expression, using our “FOIL” skills, and group the like terms.

Let’s try one!

Convert the function to standard form:

The conversion process:

22( 3) 8y x 22 12 10y x x

Unfortunately, this procedure isn’t quite as simple. However, there are 2 ways that we could proceed.◦ 1. We can complete the square◦ 2. We can make use of the vertex formula.

Let’s take a look at an example for each!

Converting from Standard Form to Vertex Form.

First, we’ll look at an “easy one”—where the value of a is 1.

We’ll work it through on the board, then I’ll reveal the steps here on the PowerPoint.

Rewrite the function in vertex form:

The conversion process: Completing the Square

2 10 12y x x

Steps in the procedure:◦ 1. take half of b◦ 2. square the result◦ 3. ADD (to complete the square) this amount to

the function and subtract this amount from the function ( aha…thus we’ve added ZERO…and not changed the value of the function)

◦ 4. Rewrite as a binomial squared, and we’re there.

◦ Half square ADD Rewrite

Completing the square (a = 1)

First, group the x terms together, leave some space.

Now, take half of the 10, square it, and add it AND subtract it.

Rewrite the first 3 terms as a binomial squared

So our vertex is at

(-5, -37)

Our solution

2 10 12y x x 2 10 12y x x

2 10 2 25125y x x 2( 5) 37y x

In this case, we first FACTOR out the value of a to get a look at a coefficient of 1 on our x squared term….it’s a bit messier, and we must be careful when we do the subtract part.

Let’s do one together.

Rewrite the function in vertex form:

A tougher challenge ( a isn’t 1)

23 24 12y x x

23 24 12y x x

We’ll do this one together

23( 8 ) 12y x x

23( 8 6 48) 121y x x

243( ) 60y x

Vertex is (4, -60)

Recall from a previous slide, that the parabola

has its vertex at:

We can use this idea to help to transform the equation into vertex form, also.

Conversion Process #2:Using the Vertex Formula

2( )f x ax bx c

,2 2

b bf

a a

Rewrite the function in vertex form, and identify the vertex.

The x coordinate for the vertex is

Let’s do an example together

22 8 10y x x

2

bx

a

2

8

a

b

8 82

2(2) 4x

Now, we plug that x value into the function, to find the y value for the vertex point.

So the vertex is (-2, -18) and the form becomes

Continuing that last example

22 8 10y x x

2x 22( 2) 8( 2) 10y

8 16 10 18y

22( 2) 18y x

Vertex Form:

◦ Vertex is at (h, k)

Standard Form:

The value of a determines stretch or compress.

The value of a determines whether the graph opens up or down.

The value of a determines the concavity of the graph.

General information slide: Quadratic Functions

2( )y a x h k

2y ax bx c

1. Change the given equation to vertex form

2. Determine the vertex. 3. Determine the y intercept. 4. Sketch the graph.

Wrap up Example

2 4 5y x x

Did you get these answers?

2( 2) 9y x Vertex Form

Vertex

Y intercept

(2, 9)

(0, 5)

Did you get these answers?

2( 2) 9y x The graph

Linear FunctionsGraphs, Slope, and

Intercepts

What is an Intercept?

Y

X

An Intercept is the coordinate where a line crosses the x or y axis

Using X and Y Intercepts to Graph a Line

The X intercept is the x coordinate (where a line crosses the x axis).

Y

X

(0,2)

(3,0)

•

•

The Y intercept is the y coordinate (where a line crosses the y axis).

Name the X intercept and the Y Intercept

Y

X

(0,2)

(3,0)

•

•

What is the value of x at the y intercept? What is the value of Y at the x-intercept?

Y

X

Graph y = 2x - 6 using x&y intercepts

1st t chart

X Y = 2x - 6

Y

X

Graph Linear Eq.


1st Make x-y table

2nd Set x = 0 and solve for y

X Y = 2x - 6

0 -6

Y

X

• (0,-6)

Graph Linear Eq.


1st Make x-y table


3rd Set y = 0 and solve for x

X Y = 2x - 6

0 -6

3 0Y

X

• (0,-6)

• (3,0)

Graph Linear Eq.


1st Make x-y table



4th Plot these 2 points and draw line

X Y = 2x - 6

0 -6

3 0Y

X

• (0,-6)

• (3,0)

Graph Linear Eq.


1st Make x-y table



4th Plot these 2 points and draw line

5th Use 3rd point to check (this is totally

optional…in fact, I don’t usually do it.)

X Y = 2x - 6

0 -6

3 0

4 2Y

X

• (0,-6)

• (3,0)

• (4,2)

Graph Linear Eq.

Graphing Horizontal & Vertical Lines

Y

X

This line has a y value of 4 for any x-value. It’s equation is

y = 4 (meaning y always equals 4)

Graphing Horizontal & Vertical Lines

Y

X

This line has a x value of 1 for any y-value. It’s equation is

x = 1 (meaning x always equals 1)

The Equation of a Vertical Line is X=Constant

Y

X

x = 1

The Equation of a Horizontal Line is Y=Constant

Y

X

y = 3

SLOPE =

Slope is a measure of STEEPNESS

RISE

RUN

The Symbol forSLOPE = m

SLOPE =

RISE

RUN

RUN

RISE•

•

How much does this line rise?

How much does it run?

(3,2)

(6,4)

(0,0) 1 2 3

1

2

4

3

5 6

4

SLOPE =

RISE

RUN

•

•

How much does this line rise?

How much does it run?

(3,2)

(6,4)

(0,0) 1 2 3

1

2

4

3

5 6

4

RUN 3

RISE 2

m=SLOPE =

•

•

(3,2)

(6,4)

(0,0) 1 2 3

1

2

4

3

5 6

4

RUN 3

RISE 2

Slopemy2 y1

x2 x1

4 2

6 3

2

3

RISE

RUNy2 y1

x2 x1

x1y1

x2y2

SLOPE mRISE

RUN

ZERO Slope --Horizontal

Positive SlopeIs Up

Negative SlopeIs Down

Undefined Slope--Vertical

Equations of a Line

There are 3 Forms of Line Equations

• Standard Form: ax+by=c

• Slope Intercept Form: y=mx+b

• Point-Slope Form y-y1=m(x-x1)

(We’ll make use of this next class)

All 3 describe the line completely

Converting fromStandard Form: ax+by=c

to Slope Intercept Form

3x 6y 12

6y 3x 12

6

6y

3

6x

12

6

y 1

2x 2 Slope Intercept Form:

y=mx+b

JUST SOLVE FOR Y

Slope Intercept Form:y=mx+b

The great thing about this form is b is the y-intercept.

This makes graphing a line incredibly easy. Check it out. If

• (0,1)

y 23 x 1

The y intercept is +1

Almost a free point on graph

Slope Intercept Form:y=mx+b

All you have to do now is use the slope to rise and run

from the intercept & connect the points.

•(0,1)

y 23 x 1

mrise

run

2

3

•

Rise 2 and Run 3 from the y-intercept & connect points.

y=mx+b when m is negative

All you have to do now is use the slope to rise and run

from the intercept & connect the points.

•(0,1)

y 23 x 1

mrise

run

2

3

•

Rise -2 and Run 3 from the y-intercept & connect points.

If the equation is not in y=mx+b form

solve for y

2

2y

5

2x

4

2

y 5

2x 2

2y 5x 4

Now it isThis line has an y intercept of -2 and rises 5 and runs 2.

Solution Steps to Solve for y:

Divide by 2

Graphing a line with slope

intercept form

2

2y

5

2x

4

2

y 5

2x 2

2y 5x 41. Solve for y:

2. Y-Intercept is 1st Point.

3. From the y-intercept

Rise 5 and run 2 for

Second Point.

4. Connect Points with line.

2y 5x 4Graph

Y

X

•(0,-2)

5{2•

y 5

2x 2 Now it is easy to graph

• (0,-2)

•

Review Steps of Graphing from the Slope Intercept Equation

1. Make sure equation is in y=mx+b form

2. Plot b(y-intercept) on graph (0,b)

3. From b, Rise and Run according to the slope to plot 2nd point.

4. Check sign of slope visually

Wrapping Up Thanks for your attention and participation.◦You have my utmost respect for working hard all day with your kids, and still hanging in there for what we’ve done!

Hang in there, only a week till Thanksgiving, and we all have much to be thankful for

If I can help in any way, don’t hesitate to shoot me an email, or give me a call.

Presented by Dr. Del Ferster. We’ll spend a bit of time looking at some “test-type” problems....

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Transcript of Presented by Dr. Del Ferster. We’ll spend a bit of time looking at some “test-type” problems....