[email protected] MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt 1 Bruce Mayer, PE Chabot...

[email protected] • MTH55_Lec-51_sec_8-3a_Quadratic_Fcn_Graphs.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§8.3 Quadratic§8.3 QuadraticFcn GraphsFcn Graphs



Review §Review §

Any QUESTIONS About• §8.2 → Quadratic Eqn Applications

Any QUESTIONS About HomeWork• §8.2 → HW-38

8.2 MTH 55



Graphs of Quadratic EqnsGraphs of Quadratic Eqns

All quadratic functions have graphs similar to y = x2. Such curves are called parabolas. They are U-shaped and symmetric with respect to a vertical line known as the parabola’s line of symmetry or axis of symmetry.

For the graph of f(x) = x2, the y-axis is the axis of symmetry. The point (0, 0) is known as the vertex of this parabola.



Example Example Graph Graph ff((xx) = 2) = 2xx22

Solution:Make T-Table andConnect-Dots

x y (x, y)

0

1

–1

2

–2

0

2

2

8

8

(0, 0)

(1, 2)

(–1, 2)

(2, 8)

(–2, 8)

x = 0 is Axis of Symm (0,0) is Vertex

x

y

(-1,2 )

(2,8)(-2,8)

-5 -4 -3 -2 -1 1 2 3 4 5

4

3

6

2

5

1

(1,2)

(0, 0)-1

-2

78



Example Example Graph Graph ff((xx) = ) = −−33xx22


Same Axis & Vertex but opens DOWNward

x y (x, y)

0

1

–1

2

–2

0

–3

–3

–12

–12

(0, 0)

(1, –3)

(–1, –3)

(2, –12)

(–2, –12)

x

y

-5 -4 -3 -2 -1 1 2 3 4 5

-3

2

-2

3

-1

1

6

54

-4

-5



Examples of Examples of axax22 Parabolas Parabolas2( )f x x

4

3

6

2

5

1

2( ) 4f x x

21( )

4f x x

2( )f x x



Graphing Graphing ff((xx) = ) = axax22

The graph of f(x) = ax2 is a parabola with • x = 0 as its axis of symmetry.

• The Origin, (0,0) as its vertex.

For Positive a the parabola opens upward For Negative a the parabola opens downward If |a| is greater than 1; e.g., 4, the parabola is

narrower (tighter) than y = x2. If |a| is between 0 and 1 e.g., ¼, the parabola

is wider (broader) than y = x2.



The Graph of The Graph of ff((xx) = ) = aa((xx – – hh))22

We could next consider graphs of f(x) = ax2 + bx + c, where b and c are not both 0.

It turns out to be convenient to first graph f(x) = a(x – h)2, where h is some constant; i.e., h is a NUMBER

This allows us to observe similarities to the graphs drawn in previous slides.



Example Example Graph Graph ff((xx) = () = (xx−2−2))22


The Vertex SHIFTED 2-Units to the Right

x y (x, y)

0

1

–1

2

3

4

4

1

9

0

1

4

(0, 4)

(1, 1)

(–1, 9)

(2, 0)

(3, 1)

(4, 4)

x

y

-5 -4 -3 -2 -1 1 2 3 4 5

4

3

6

2

5

1

-1

-2

78

vertex



Graphing Graphing ff((xx) = ) = aa((xx−−hh))22

The graph of y = f(x) = a(x – h)2 has the same shape as the graph of y = ax2.

If h is positive, the graph of y = ax2 is shifted h units to the right.

If h is negative, the graph of y = ax2 is shifted |h| units to the left.

The vertex is (h, 0) and the axis of symmetry is x = h.



Graph of Graph of ff((xx) = ) = aa((xx – – hh))22 + + kk

Given a graph of f(x) = a(x – h)2, what happens if we add a constant k?

Suppose we add k = 3. This increases f(x) by 3, so the curve moves up • If k is negative, the curve moves down.

The axis of symmetry for the parabola remains x = h, but the vertex will be at (h, k), or equivalently (h, f(h))• f(h) = a([h] – h)2 + k = 0 + k → f(h) = k



Example Example Graph Graph

The Vertex SHIFTED 3-Units Left and1-Unit Down

Make T-Table andConnect-Dots

x y (x, y)

0

–1

–2

–3

–4

–5

-11/2

–3

–3/2

–1

–3/2

–3

(0, -11/2)

(–1, –3)

(–2, –3/2)

(–3, –1)

(–4, –3/2)

(–5, –3)

21( ) ( 3) 1.

2f x x

x

y

-5 -4 -3 -2 -1 1 2 3 4 5

-3

2

-2

3

-1

1

-4

-5-6

-7

-8vertex



Quadratic Fcn in Standard FormQuadratic Fcn in Standard Form

The Quadratic Function Written in STANDARD Form:

f x a x h 2 k , a 0

The graph of f is a parabola with vertex (h, k). The parabola is symmetric with respect to the line x = h, called the axis of the parabola. If a > 0, the parabola opens up, and if a < 0, the parabola opens down.



Example Example Find Quadratic Fcn Find Quadratic Fcn

Find the standard form of the quadratic function whose graph has vertex (−3, 4) and passes through the point ( −4, 7).

SOLUTION: Let y = f(x) be the quadratic function. Then

y a x h 2 k

y a x 3 2 4

y a x 3 2 4

7 a –4 3 2 4

a 3

Hence,

y 3 x 3 2 4



Graphing Graphing ff((xx) = ) = aa((xx – – hh))22 + + kk

1. The graph is a parabola. • Identify a, h, and k

2. Determine how the parabola opens. • If a > 0 (positive), the parabola opens up.

• If a < 0 (negative), the parabola opens down.

3. Find the vertex. The vertex is (h, k). • If a > 0 (or a < 0), the function f has a

minimum (or a maximum) value k at x = h




4. Find the x-intercepts.• Find the x-intercepts (if any) by setting

f(x) = 0 and solving the equation a(x – h)2 + k = 0 for x.

– Solve by one of: [Factoring + ZeroProducts], Quadratic Formula, CompleteSquare

– If the solutions are real numbers, they are the x-intercepts.

– If the solutions are NOT Real Numbers, the parabola either lies above the x–axis (when a > 0) or below the x–axis (when a < 0).




5. Find the y-intercept• Find the y-intercept by replacing x with 0.

Then y = f(0) = ah2 + k is the y-intercept.

6. Sketch the graph• Plot the points found in Steps 3-5 and join

them by a parabola. – If desired, show the axis of symmetry, x = h,

for the parabola by drawing a dashed vertical line



Example Example Graph Graph f x 2 x 3 2 8.

SOLUTION

Step 1 a = 2, h = 3, and k = –8Step 2 a = 2, a > 0, the parabola opens up.Step 3 (h, k) = (3, –8); the function f has a

minimum value –8 at x = 3.Step 4 Set f (x) = 0 and solve for x.

0 2 x 3 2 8

8 2 x 3 2

4 x 3 2

x 3 2

x 5 or x 1

x-intercepts: 1 and 5




SOLUTION cont.Step 5 Replace x with 0.

f 0 2 0 3 2 8

2 9 8 10

y-intercept is 10 .

Step 6 axis: x = 3, opens up, vertex: (3, –8), passes through (1, 0), (5, 0) and (0, 10), the graph is y = 2x2 shifted three units right and eight units down.




SOLUTION cont.• Sketch Graph

Using the 4 points– Vertex

– Two x-Intercepts

– One y-Intercept



Completing the SquareCompleting the Square

By completing the square, we can rewrite any polynomial ax2 + bx + c in the form a(x – h)2 + k.

Once that has been done, the procedures just discussed enable us to graph any quadratic function.



Example Example Graph Graph 2( ) 2 1.f x x x

SOLUTION

f (x) = x2 – 2x – 1

= (x2 – 2x) – 1

= (x2 – 2x + 1) – 1 – 1

= (x2 – 2x + 1 – 1) – 1

= (x – 1)2 – 2

The vertex is at (1, −2) The Parabola

Opens UP

x

y

-5 -4 -3 -2 -1 1 2 3 4 5

-3

2

-2

3

-1

1

6

54

-4

-5

Adding ZERO



Example Example Graph Graph SOLUTION

The vertex is at (3/2, 3/2)

2( ) 2 6 3.f x x x

f (x) = –2x2 + 6x – 3 = –2(x2 – 3x) – 3

= –2(x2 – 3x + 9/4) – 3 + 18/4

= –2(x2 – 3x + 9/4 – 9/4) – 3

= –2(x – 3/2)2 + 3/2

x

y

-5 -4 -3 -2 -1 1 2 3 4 5

-3

2

-2

3

-1

1

6

54

-4

-5

CompleteSquare



The Vertex of a ParabolaThe Vertex of a Parabola

By the Process of Completing-the-Square we arrive at a FORMULA for the vertex of a parabola given by f(x) = ax2 + bx + c:

24, or , .

2 2 2 4

b b b ac bf

a a a a

• The x-coordinate of the vertex is −b/(2a).

• The axis of symmetry is x = −b/(2a).

• The second coordinate of the vertex is most commonly found by computing f(−b/[2a])



Graphing Graphing ff((xx) = ) = axax22 + bx + c + bx + c

1. The graph is a parabola. Identify a, b, and c

2. Determine how the parabola opens• If a > 0, the parabola opens up.

• If a < 0, the parabola opens down

3. Find the vertex (h, k). Use the formula

h, k b

2a, f

b

2a

.




4. Find the x-interceptsLet y = f(x) = 0. Find x by solving the equation ax2 + bx + c = 0.

• If the solutions are real numbers, they are the x-intercepts.

• If not, the parabola either lies – above the x–axis when a > 0

– below the x–axis when a < 0




5. Find the y-intercept. Let x = 0. The result f(0) = c is the y-intercept.

6. The parabola is symmetric with respect to its axis, x = −b/(2a)

• Use this symmetry to find additional points.

7. Draw a parabola through the points found in Steps 3-6.




f x 2x2 8x 5.

Step 1 a = –2, b = 8, and c = –5Step 2 a = –2, a < 0, the parabola opens down.Step 3 Find (h, k).

h b

2a

8

2 2 2

k f 2 2 2 2 8 2 5 3

h, k 2, 3 Maximum value of y = 3 at x = 2




f x 2x2 8x 5.

Step 4 Let f (x) = 0.

f 0 2 0 2 8 0 5

y-intercept is 5 .Step 5 Let x = 0.

2x2 8x 5 0

x 8 8 2 4 2 5

2 2 4 6

2

x-intercepts are 4 6

2 and

4 6

2.




f x 2x2 8x 5.

Step 6 Axis of symmetry is x = 2. Let x = 1, then the point (1, 1) is on the graph, the symmetric image of (1, 1) with respect to the axis x = 2 is (3, 1). The symmetric image of the y–intercept (0, –5) with respect to the axis x = 2 is (4, –5).

Step 7 The parabola passing through the points found in Steps 3–6 is sketched on the next slide.



Example Example Graph Graph f x 2x2 8x 5.

SOLUTION cont.• Sketch Graph

Using the pointsJust Determined

f x 2x2 8x 5



Find Domain & RangeFind Domain & Range Given the graph of

f(x) = −2x2 +8x − 5

Find the domain and range for f(x)

SOLUTION Examine the Graph to find that the:• Domain is (−∞, ∞)

• Range is (−∞, 3]



WhiteBoard WorkWhiteBoard Work

Problems From §8.3 Exercise Set• 4, 16, 22, 30

The Directrix of a Parabola• A line perpendicular to the axis

of symmetry used in the definition of a parabola. A parabola is defined as follows: For a given point, called the focus, and a given line not through the focus, called the directrix, a parabola is the locus of points such that the distance to the focus equals the distance to the directrix.



All Done for TodayAll Done for Today

GeometricComplete

TheSquare

22

2

52510

10

xxx

xx



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

–

srsrsr 22



Graph Graph yy = | = |xx||

Make T-tablex y = |x |

-6 6-5 5-4 4-3 3-2 2-1 10 01 12 23 34 45 56 6

x

y

-6

-5

-4

-3

-2

-1

0

1

2

3

4

5

6

-6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6

file =XY_Plot_0211.xls



-3

-2

-1

0

1

2

3

4

5

-3 -2 -1 0 1 2 3 4 5

M55_§JBerland_Graphs_0806.xls -5

-4

-3

-2

-1

0

1

2

3

4

5

-10 -8 -6 -4 -2 0 2 4 6 8 10

M55_§JBerland_Graphs_0806.xls

x

y

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