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Finding the Vertex of Quadratic Functions, Lecture notes of Advanced Physics

Instructions and examples on how to find the vertex of a quadratic function, which is the point where the parabola reaches its maximum or minimum value. The vertex can be found by using the formula x = โˆ’b/2a, where a and b are coefficients of the quadratic equation ax^2 + bx + c. The document also explains how to find the y-coordinate of the vertex by substituting the x-coordinate back into the quadratic equation.

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16-week Lesson 23 (8-week Lesson 19) The Vertex of a Quadratic Function
1
As shown in the previous set of notes, the vertex of a quadratic function
provides information about many features of that function and its graph
(increasing/decreasing intervals, line of symmetry, maximum/minimum
function value, range). However when a quadratic function is expressed in
polynomial form (๐‘“(๐‘ฅ)= ๐‘Ž๐‘ฅ2+๐‘๐‘ฅ +๐‘), the vertex of the quadratic
function is not obvious. One way to find the vertex of a quadratic function
that is in polynomial form is to use the formula ๐‘ฅ = โˆ’๐‘
2๐‘Ž to find the
๐‘ฅ-coordinate of the vertex. Once you have the ๐‘ฅ-coordinate, you can find
the ๐‘ฆ-coordinate by replacing ๐‘ฅ with โˆ’๐‘
2๐‘Ž in the quadratic function
(๐‘“(โˆ’๐‘
2๐‘Ž)). Just like the quadratic formula that weโ€™ve worked with before,
the formula ๐‘ฅ = โˆ’๐‘
2๐‘Ž is also derived by completing the square (more on this
later).
Example 1: Find the vertex of the quadratic function
๐‘“(๐‘ฅ)= โˆ’2๐‘ฅ2+16๐‘ฅ โˆ’ 26.
๐‘ฅ = โˆ’๐‘
2๐‘Ž
๐‘ฅ = โˆ’(16)
2(โˆ’2)
๐‘ฅ = โˆ’16
โˆ’4
๐‘ฅ = 4
๐‘“(4)= โˆ’2(4)2+16(4)โˆ’26
๐‘“(4)= โˆ’32+64โˆ’26
๐‘“(4)= 6
๐•๐ž๐ซ๐ญ๐ž๐ฑ:(๐Ÿ’,๐Ÿ”)
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Download Finding the Vertex of Quadratic Functions and more Lecture notes Advanced Physics in PDF only on Docsity!

As shown in the previous set of notes, the vertex of a quadratic function

provides information about many features of that function and its graph

(increasing/decreasing intervals, line of symmetry, maximum/minimum

function value, range). However when a quadratic function is expressed in

polynomial form

2

, the vertex of the quadratic

function is not obvious. One way to find the vertex of a quadratic function

that is in polynomial form is to use the formula ๐‘ฅ =

โˆ’๐‘

2 ๐‘Ž

to find the

๐‘ฅ-coordinate of the vertex. Once you have the ๐‘ฅ-coordinate, you can find

the ๐‘ฆ-coordinate by replacing ๐‘ฅ with

โˆ’๐‘

2 ๐‘Ž

in the quadratic function

โˆ’๐‘

2 ๐‘Ž

)). Just like the quadratic formula that weโ€™ve worked with before,

the formula ๐‘ฅ =

โˆ’๐‘

2 ๐‘Ž

is also derived by completing the square (more on this

later).

Example 1 : Find the vertex of the quadratic function

2

2

Example 2 : Find the vertex of the following quadratic functions.

a. ๐‘“(๐‘ฅ) = 5 ๐‘ฅ

2

  • 20 ๐‘ฅ + 14 b. ๐‘”(๐‘ฅ) = 9 ๐‘ฅ

2

b.

c. โ„Ž

2

โˆ’ 6 ๐‘ฅ โˆ’ 5 d. ๐‘—

3

4

2

All the quadratic functions weโ€™ve seen so far have been in polynomial

form

2

. However if a quadratic function is not

given in polynomial form, it can be converted to polynomial form by

using polynomial multiplication and combining like terms. This is what

weโ€™ll do on the next example.

๐‘ฅ =

โˆ’๐‘

2 ๐‘Ž

๐‘” (

4

3

) = 9 (

4

3

)

2

โˆ’ 24 (

4

3

) + 16

๐‘ฅ =

โˆ’

( โˆ’ 24

)

2 ( 9 )

๐‘” (

4

3

) = 9 (

16

9

) โˆ’ 32 + 16

๐‘ฅ =

24

18

๐‘” (

4

3

) = 16 โˆ’ 32 + 16

๐‘ฅ =

4

3

๐‘” (

4

3

) = 0

๐‘ฅ =

4

3

๐Ÿ’

๐Ÿ‘

๐‘ฅ =

โˆ’๐‘

2 ๐‘Ž

โ„Ž(โˆ’ 1 ) = โˆ’ 3 (โˆ’ 1 )

2

โˆ’ 6 (โˆ’ 1 ) โˆ’ 5

๐‘ฅ =

โˆ’

( โˆ’ 6

)

2 (โˆ’ 3 )

โ„Ž

( โˆ’ 1

) = โˆ’ 3

( 1

)

  • 6 โˆ’ 5

๐‘ฅ =

6

โˆ’ 6

โ„Ž(โˆ’ 1 ) = โˆ’ 3 + 6 โˆ’ 5

๐‘ฅ = โˆ’ 1 โ„Ž(โˆ’ 1 ) = โˆ’ 2

๐‘ฅ =

4

3

Example 4 : Find the vertex of the following quadratic functions, and state

whether the vertex is the maximum or minimum point on the parabola.

a. ๐‘“

2

โˆ’ 12 ๐‘ฅ + 21 b. ๐‘š

โˆ’๐‘

2 ๐‘Ž

2

โˆ’

( โˆ’ 12

)

2

( 1

)

2

12

2

โˆ’๐‘

2 ๐‘Ž

โˆ’

( โˆ’ 12

)

2

( โˆ’ 3

)

2

12

โˆ’ 6

Example 5 : Find the vertex and the zeros for each of the following

quadratic functions.

a. ๐‘˜

b. ๐‘”

2

2

2

โˆ’

( 6

)

2

( โˆ’ 1

)

โˆ’

( โˆ’ 6

)

2

( โˆ’ 3

)

โˆ’ 6

โˆ’ 2

6

โˆ’ 6

2

Since ๐‘”

( ๐‘ฅ

) is already in polynomial form,

I do not need to change it in any way to

find the vertex, like I have to do with ๐‘˜(๐‘ฅ)

in part a.

Example 7: Find the vertex and the range for each of the following

quadratic functions.

a. ๐‘˜

b. ๐‘”

2

2

โˆ’ 2

2

( 1

)

2

โˆ’ 2

2

โˆ’(โˆ’ 8 )

2

( โˆ’ 2

)

8

โˆ’ 4

2

In order to find the range of each function, I simply need to use the ๐‘ฆ-

coordinate of each vertex and determine whether that is the largest

output or the smallest output. To do so, I use the leading coefficient

of each function to determine whether the graph is opening up or

down.

The leading coefficient of The leading coefficient of

๐‘˜(๐‘ฅ) is โˆ’ 2 , so the graph of ๐‘”(๐‘ฅ) is 1 , so the graph of ๐‘”(๐‘ฅ)

๐‘˜(๐‘ฅ) is opening down. That is opening up. That means

means 50 is the largest output โˆ’ 9 is the smallest output

]

[

Example 8: Find the vertex and the increasing/decreasing intervals for

each of the following quadratic functions.

a. ๐‘“(๐‘ฅ) = โˆ’

1

2

( 3 โˆ’ ๐‘ฅ)(๐‘ฅ + 2 ) b. โ„Ž(๐‘ฅ) =

2

5

2

12

5

23

5

1

2

2

โˆ’(โˆ’

12

5

)

2 (

2

5

)

1

2

2

1

2

12

5

4

5

โˆ’(โˆ’

1

2

)

2 (

1

2

)

12

4

1

2

1

1

2

2

5

2

12

5

23

5

1

2

1

2

1

2

1

2

2

5

36

5

23

5

1

2

5

2

5

2

18

5

36

5

23

5

25

8

๐Ÿ

๐Ÿ

๐Ÿ๐Ÿ“

๐Ÿ–