• WARMUP

(3,2) X=3

All Real Numbers

y ≤ 2

Reflected over x

Stretched vertically, right 3, up 2

(2,0) (4,0)(0 , -16)

< x < 3𝟑<𝒙<∞

VERTEX FORM: STANDARD FORM:

SKILL : CONVERT BETWEEN THE TWO Algebraic and Graphical Conversion

Vertex to Standard form – Use the MATHStandard to Vertex form – Use the Graphing Calculator

PRACTICE: Calculate/Break Down

FOIL!!First, Outer, Inner, Last

A third way to do this!!Combine like terms

x2 + 8x + 16

Distribute first factor into second factor

– 6x + 9 + 9

x 4

x

4

x2 4x

4x 16

No matter which method you choose, put original equation into y1 and your answer into y2 and check to be sure the tables match!!!

Finally, a FOURTH way to do this, using the calculator.1) Put the equation you are given into the calculator, in y=2) Look at the table, and pick a few points.

Preferably, pick at least one point on either side of the vertex.3) Go to STAT Edit, and enter the points you chose into L1 (x-values) and L2 (y-values).4) Go to STAT Calc 5 and get your equation.5) As always, put both equations into y= and make sure that the tables match.

What does the graph look like?We know that the vertex is at (2,4) because it’s in vertex form.We know that the graph opens up, because it’s positiveSo, the graph looks roughly like this:

(𝒙−𝟐 ) (𝒙 −𝟐 )+𝟒FOIL METHOD

First, Outer, Inner, Last

DISTRIBUTE

x -2

x

-2

x2 -2x

-2x 4BOX SAME ANSWER EVERY TIME!

𝒙𝟐−𝟒 𝒙+𝟖(𝟐 ,𝟒) 𝒙=𝟐

𝑨𝒍𝒍 𝑹𝒆𝒂𝒍𝑵𝒖𝒎𝒃𝒆𝒓𝒔𝒚 ≥𝟒

𝑴𝒊𝒏𝒊𝒎𝒖𝒎𝑵𝒐𝒏𝒆(𝟎 ,𝟖)𝟐<𝒙<∞−∞<𝒙<𝟐

-2 FOIL METHOD-2First, Outer, Inner, Last

1) 1)

DISTRIBUTE) )

x -1

x

-1

x2 -x

-x 1BOX

−𝟐 𝒙𝟐+𝟒 𝒙−𝟐(𝟏 ,𝟎) 𝒙=𝟏

𝑨𝒍𝒍 𝑹𝒆𝒂𝒍𝑵𝒖𝒎𝒃𝒆𝒓𝒔𝒚 ≤𝟎

𝑴𝒂𝒙𝒊𝒎𝒖𝒎

(𝟎 ,−𝟐)

𝟏<𝒙<∞−∞<𝒙<𝟏

(𝟏 ,𝟎)

This is already in standard formPut it into y=, and use 2nd Trace 4 to find the

vertex.The vertex is at (3,2)

We can also find the vertex by hand.The x-value of the vertex is found by

Once the x-value has been found, use it to find y.

If x = 3, then

Once again, the vertex is at (3,2)

(𝟑 ,𝟐) 𝒙=𝟑𝑹𝒆𝒇𝒍𝒆𝒄𝒕 𝒐𝒗𝒆𝒓 𝒙 ;

𝑽𝒆𝒓𝒕𝒊𝒄𝒂𝒍 𝒔𝒕𝒓𝒆𝒕𝒄𝒉𝒃𝒚 𝟐 ,𝒓𝒊𝒈𝒉𝒕 𝟑 ,𝒖𝒑𝟐

𝒚=−𝟐 (𝒙−𝟑 )𝟐+𝟐𝑨𝒍𝒍 𝑹𝒆𝒂𝒍𝑵𝒖𝒎𝒃𝒆𝒓𝒔𝒚 ≤𝟐

𝑴𝒂𝒙𝒊𝒎𝒖𝒎

(𝟎 ,−𝟏𝟔)

𝟑<𝒙<∞−∞<𝒙<𝟑

(𝟐 ,𝟎 )(𝟒 ,𝟎)

−∞−∞

(𝟐𝟑 ,

𝟒𝟑 ) 𝒙=

𝟐𝟑

𝑹𝒆𝒇𝒍𝒆𝒄𝒕 𝒐𝒗𝒆𝒓 𝒙 ;𝑽𝒆𝒓𝒕𝒊𝒄𝒂𝒍 𝒔𝒕𝒓𝒆𝒕𝒄𝒉𝒃𝒚 𝟑 ,𝒓𝒊𝒈𝒉𝒕 𝟐𝟑 ,𝒖𝒑

𝟒𝟑

𝒚=−𝟑(𝒙−𝟐𝟑 )𝟐+𝟒𝟑

𝑨𝒍𝒍 𝑹𝒆𝒂𝒍𝑵𝒖𝒎𝒃𝒆𝒓𝒔𝒚 ≤𝟒𝟑

𝑴𝒂𝒙𝒊𝒎𝒖𝒎

(𝟎 ,𝟎)

𝟐𝟑<𝒙<∞

−∞<𝒙<𝟐𝟑

(𝟎 ,𝟎 )(𝟒𝟑 ,𝟎)

−∞−∞

Find Parabola Fitting over Points

Estimation/ModelingQuadratic Regression and Calculations

1 -7 5

𝑥2−7 𝑥+5

-1 4 -7

−𝑥2+4 𝑥−7

2 -7 8

2 𝑥2−7 𝑥+8

-17.07 110.73 -3

−17.07 𝑥2+110.73 𝑥−3Enter equation in y= and look at the table for x = 6.

OR – simply plug 6 into the equation by hand.

Either way, we get that the ball will be around 47 feet off the ground after 6 seconds.

Use the three points that we have to find the equation, then use the equation to find the two missing y-values.

Put 1, 2, and 3 into L1 and 25, 38, and 75 into L2STAT Calc 5 gets us the equation of

12 -23 36

12𝑥2−23 𝑥+36

Put into y=, then go 2nd Graph and look at the table for x=4 and x=5.

136 221

-.1 1 6

− .1 𝑥2+𝑥+6To find x = 8.5 on the graph, go to 2nd Window, and change TblStart= to 8.5.2nd Graph will show you 7.275

-.1 1 6

− .1 𝑥2+𝑥+6You can also use the STO button to find this. Type 8.5, STO, x. Then type in the equation (not in y=) to get your answer.Again, you will get 7.275

FIND THE VERTEX!! The y-value will be the lowest point, and the x-value will be the distance from the tower.To use the calculator, you have to play with your window to see the graph. Remember, the y-intercept is at 150. I used a range of 0 to 100 on the x-axis and from 0 to 165 on the y.By hand, use to find x, then plug that into the equation to find y.Either way, the vertex is (35,27.5)The cable gets within 27.5 feet above the roadway, and is 35 feet from the tower at that point.