Section 3.1: Quadratic Functions 1

MAT121: SECTION 3.1

QUADRATIC FUNCTIONS

ANATOMY OF A QUADRATIC FUNCTION: GRAPH

• Shape: Parabola – it may open upwards or down.

• Vertex: Represented as (ℎ, 𝑘). It can be the maximum or minimum point depending on how the

parabola opens.

• Axis of Symmetry: Pass through the vertex. Always represented as the equation, 𝑥 = ℎ.

Quadratic functions can be expressed in two ways:

• Polynomial Form: 𝑓 𝑥 = 𝑎𝑥! + 𝑏𝑥 + 𝑐

• Vertex Form: 𝑓 𝑥 = 𝑎 𝑥 − ℎ ! + 𝑘

VERTEX FORM

𝑓 𝑥 = 𝑎 𝑥 − ℎ ! + 𝑘 Note its similarity to the transformations from Section 2.6

• The sign of a determines the direction in which the parabola opens and whether the vertex is a

maximum or minimum point.

o a > 0 (positive) then the parabola opens upward and the vertex is a minimum.

o a < 0 (negative) then the parabola opens downward and the vertex is a maximum.

• In this form, the vertex is easily determined. (ℎ, 𝑘) are the coordinates of the vertex.

Section 3.1: Quadratic Functions 2

Consider 𝑓 𝑥 = 𝑥 − 1 ! + 2

• Does the parabola open up or down? Why? Up, because a is positive.

• State the vertex. (1, 2)

• Is this a max or min? Minimum

• State the Axis of Symmetry. 𝑥 = 1

• State the Domain and Range. 𝐷𝑜𝑚𝑎𝑖𝑛: −∞,∞ 𝑅𝑎𝑛𝑔𝑒: [2,∞)

Consider 𝑓 𝑥 = 𝑥 + 4 ! + 1.

The function can be rewritten to fit the model. 𝑓 𝑥 = 𝑥 − −4 ! + 1

• State the vertex. (−4, 1)

• Is this a max or min? Minimum

• State the Axis of Symmetry. 𝑥 = −4

• Graph the function. What are the x- and y-intercepts?

x-intercepts: Consider the vertex. It’s above the x-axis and is the minimum, so no x-intercepts.

y-intercept: 𝑓 0 = 0+ 4 ! + 1 = 17 so the y-intercept is (0, 17)

ON YOUR OWN

Consider 𝑓 𝑥 = −2 𝑥 + 6 ! − 5.

• Does the parabola open up or down? Down

• State the vertex. Is it a max or min? Max

• State the y-intercept. (0,−77)

• State the Domain and Range. 𝐷: −∞,∞ 𝑅: (−∞,−5]

• Graph it.

Section 3.1: Quadratic Functions 3

POLYNOMIAL FORM

𝑓 𝑥 = 𝑎𝑥! + 𝑏𝑥 + 𝑐

• The sign of a determines the direction in which the parabola opens and whether the vertex is a

maximum or minimum point.

Consider 𝑓 𝑥 = 3𝑥! − 6𝑥 + 1.

• Does the parabola open up or down? Up, because the coefficient on x2 is positive.

• State the y-intercept. 𝑓 0 = 1, so the y-intercept is (0,1).

• Convert the function to vertex form by completing the square.

3𝑥! − 6𝑥 + 1

3(𝑥! − 2𝑥 )+ 1

3 𝑥! − 2𝑥 + 1 + 1− (3 ∙ 1)

3 𝑥 − 1 ! − 2

• State the vertex. (1,−2) It’s a minimum.

• Find the x-intercepts. Recall (𝑥, 0) represents the x-intercept.

Using the vertex form.

3 𝑥 − 1 ! − 2 = 0

3 𝑥 − 1 ! = 2

𝑥 − 1 ! =23

𝑥 − 1 = ±23

𝑥 = 1±23

Using the polynomial form and DQUAD.

3𝑥! − 6𝑥 + 1

𝑎 = 3

𝑏 = −6

𝑐 = 1

𝑥 = − − 6 ± 24

6

𝑥 = 6 ± 2 6

6 = 3± 63

Section 3.1: Quadratic Functions 4

FINDING AN EXPRESSION FOR VERTEX

Converting the polynomial model to the vertex model yields (page 323)

𝑓 𝑥 = 𝑎 𝑥 +𝑏2𝑎

!

+ 𝑐 −𝑏!

4𝑎

where ℎ!!!""#$%&'() !" !"#$"% = − !!!

and 𝑘!!!""#$%&'() !" !"#$"% = 𝑐 − !!

!!

Make it easier!

Given 𝑓 𝑥 = 𝑎𝑥! + 𝑏𝑥 + 𝑐, the x-coordinate of the vertex is found from – !!!

.

The y-coordinate of the vertex is found from 𝑓 – !!!

.

Consider 𝑓 𝑥 = 2𝑥! + 8𝑥 + 3. Find the vertex.

ℎ = 𝑥 𝑐𝑜𝑜𝑟𝑑𝑖𝑎𝑛𝑡𝑒 𝑜𝑓 𝑉𝑒𝑟𝑡𝑒𝑥 = −8

2 2 = −2

𝑘 = 𝑦 𝑐𝑜𝑜𝑟𝑑𝑖𝑎𝑛𝑡𝑒 𝑜𝑓 𝑉𝑒𝑟𝑡𝑒𝑥 = 𝑓 −2 = 2 −2 ! + 8 −2 + 3 = −5

Thus the vertex is (−2,−5).

ON YOUR OWN

Find the vertex of 𝑓 𝑥 = −𝑥! − 6𝑥 + 2.

ℎ = 𝑥 𝑐𝑜𝑜𝑟𝑑𝑖𝑎𝑛𝑡𝑒 𝑜𝑓 𝑉𝑒𝑟𝑡𝑒𝑥 = 6

2 −1 = −3

𝑘 = 𝑦 𝑐𝑜𝑜𝑟𝑑𝑖𝑎𝑛𝑡𝑒 𝑜𝑓 𝑉𝑒𝑟𝑡𝑒𝑥 = 𝑓 −3 = − −3 ! − 6 −3 + 2 = 11

Thus the vertex is (−3, 11).

Section 3.1: Quadratic Functions 5

(−3, 11)

GRAPHING A QUADRATIC

1. Determine if parabola opens up or down.

2. Find the vertex.

3. Find all intercepts.

4. State the Domain and Range.

5. Sketch the graph.

Consider 𝑓 𝑥 = −𝑥! − 4𝑥 − 7.

1. Opens down.

2. Vertex: ℎ = − !!! !!

= −2

𝑘 = 𝑓 −2 = − −2 ! − 4 −2 − 7 = −3

(−2,−3) Maximum

3. x-intercept: Via DQUAD – none

y-intercept: (0,−7)

4. Domain: −∞,∞ Range: (−∞,−3]

5.

Section 3.1: Quadratic Functions 6

APPLICATIONS

Suppose that a family wants to fence in an area of their yard for a vegetable garden to keep out deer.

One side is already fenced from the neighbor’s property.

• If the family has enough money to buy 160 feet of fencing, what dimensions would produce the

maximum area for the garden?

• What is the maximum area?

1. Start with what you know. 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟 = 2𝑥 + 𝑦 = 160

2. What is being maximized? 𝐴𝑟𝑒𝑎 = 𝑥𝑦

3. Solve perimeter for y and substitute into area to create the quadratic expression.

2𝑥 + 𝑦 = 160

𝑦 = −2𝑥 + 160

𝐴𝑟𝑒𝑎 = 𝑥𝑦 = 𝑥 −2𝑥 + 160 = −2𝑥! + 160𝑥

4. Since the vertex is the maximum point of this area function, find the x-coordinate of the vertex.

𝑥 = −1602 −2 = 40 𝑓𝑒𝑒𝑡

NOTE: Substituting the x-value into the quadratic expression yields the maximum area not the y-dimension.

To find the y-dimension of the garden, substitute x into 𝑦 = −2𝑥 + 160

𝑦 = −2 40 + 160 = 80 𝑓𝑒𝑒𝑡

5. Find the maximum area.

𝐴𝑟𝑒𝑎 = 𝑥𝑦 = 40 80 = 3200 𝑠𝑞 𝑓𝑒𝑒𝑡

Section 3.1: Quadratic Functions 7

In the shot put, an athlete throws the shot at an angle of 65°. Its height, 𝑔(𝑥), in feet, can be modeled

𝑔 𝑥 = −0.04𝑥! + 2.1𝑥 + 6.1

where x is the shot’s horizontal distance, in feet, from its point of release.

• What is the horizontal distance of the shot at which the maximum height occurs?

• What is the maximum height?

• What is the shot’s maximum horizontal distance (distance of the throw)?

• At what distance does the shot reach 5 feet in height?

1. The horizontal distance of the maximum height is the x-coordinate of the vertex.

𝑥 = −2.1

2 −0.04 = 26.25 𝑓𝑒𝑒𝑡

2. The maximum height is found by

𝑔 26.25 = 33.66 𝑓𝑒𝑒𝑡

3. To find the distance of the throw, calculate the x-intercepts.

Via DQUAD, 𝑥 = −2.76 𝑎𝑛𝑑 55.26. Choose the one that makes

sense. 𝑥 = 55.26 𝑓𝑒𝑒𝑡

4. We’ve been given a y-value and asked to find the x-value.

5 = −0.04𝑥! + 2.1𝑥 + 6.1

0 = −0.04𝑥! + 2.1𝑥 + 1.1

Via DQUAD, 𝑥 = 53.02 𝑓𝑒𝑒𝑡