1 Topic 7.4.3 Quadratic Graphs and Completing the Square.

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Topic 7.4.3Topic 7.4.3

Quadratic Graphs and Completing the SquareQuadratic Graphs and Completing the Square

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Topic7.4.3


California Standards:21.0: Students graph quadratic functions and know that their roots are the x-intercepts.

22.0: Students use the quadratic formula or factoring techniques or both to determine whether the graph of a quadratic function will intersect the x-axis in zero, one, or two points.

What it means for you:You’ll graph quadratic functions by first completing the square of the equation.

Key words:• quadratic• completing the square• parabola• intercept• vertex• line of symmetry• root

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Topic7.4.3


If there are no x-intercepts, then it’s impossible to find the vertex by saying that the vertex is halfway between the x-intercepts (like you saw in Topic 7.4.2).

But you can use the method of completing the square.

This means writing your equation in the form y = (x + k)2 + p.

Graph doesn’t cross the x-axis.

4Solution follows…

Topic7.4.3


Solution

Sketch the graph of y = x2 – 6x + 10.

You could try to find the x-intercepts by factoring the equation:

x2 – 6x + 10 = 0

This time, the left-hand side doesn’t factor.

Example 1

Solution continues…

So to find the solutions you could try the quadratic formula with a = 1, b = –6, and c = 10.

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Topic7.4.3


Solution (continued)


Example 1


Substitute in a = 1, b = –6 and c = 10

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Topic7.4.3




However, since you cannot take the square root of a negative number, this tells you that the quadratic function has no real roots — the equation can’t be solved using real numbers.

This means that the graph of y = x2 – 6x + 10 never crosses the x-axis.

Example 1

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Topic7.4.3


But this doesn’t mean that you can’t find the vertex — you just have to use a different method.

The trick is to write the equation of the quadratic in the form y = (x + k)2 + p — you need to complete the square.

So take the first two terms of the quadratic, and add a number to make a perfect square.

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Topic7.4.3




x2 – 6x + 10 = (x – 3)2 + 1

Therefore the function you need to sketch is y = (x – 3)2 + 1.

Example 1


Write the equation y = x2 – 6x + 10 in the form y = (x + k)2 + p.

x2 – 6x + = (x – 3)2

x2 – 6x + 10 = (x – 3)2 – + 10

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Topic7.4.3




The minimum value that (x – 3)2 takes is 0 (since a squared number cannot be negative).

Therefore the minimum value of y = (x – 3)2 + 1 is 0 + 1 = 1.

This minimum value occurs at x = 3 (the value for x where (x – 3)2 = 0).So the coordinates of the vertex of the parabola are (3, 1).

Example 1

As before, the line of symmetry passes through the vertex — so the line of symmetry is x = 3.

Find the coordinates of the vertex of the graph of y = (x – 3)2 + 1.

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Topic7.4.3


The graph of the quadratic function y = (x + k)2 + p has its vertex at (–k, p).

The line of symmetry of the graph is x = –k.

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Topic7.4.3



The graph of y = x2 – 6x + 10 = (x – 3)2 + 1:

Example 1


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4x2 – 12x + 11

Solution follows…

Topic7.4.3


Solution

Complete the square for 4x2 – 12x + 11. Then find the vertex and line of symmetry of y = 4x2 – 12x + 11.

This is a concave-up parabola, since the coefficient of x2 is positive.

Example 2


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Topic7.4.3



Complete the square for 4x2 – 12x + 11. Then find the vertex and line of symmetry of y = 4x2 – 12x + 11.

Example 2

Since the minimum value of is 0, the minimum value

of 4 + 2 must be 0 + 2 = 2. This minimum value

occurs at x = .3 2

and the line of symmetry is x = .

So the vertex of the graph of y = 4x2 – 12x + 11 is at , 3 2

As before, put x = 0 to find the y-intercept — this is at y = 11.

Find the coordinates of the minimum point of the graph of y = 4 x – + 2.32

2

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Topic7.4.3

Example 3

Solution follows…


Write 4x – x2 – 7 in the form a(x + k)2 + m, and sketch the graph.

Solution

4x – x2 – 7 = –x2 + 4x – 7

= –[x2 – 4x + 7]

= –[(x – 2)2 + 3]

= –(x – 2)2 – 3

But (x – 2)2 is never negative — the minimum value it takes is 0.

So –(x – 2)2 can never be positive, and the maximum value it can take is 0.


Factor out –1 to make completing the square easier.

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Topic7.4.3

Example 3


Write 4x – x2 – 7 in the form a(x + k)2 + m, and sketch the graph.


This means that the maximum value of –(x – 2)2 – 3 must be –3, which it takes when x – 2 = 0 — that is, at x = 2.

So the vertex of the graph is at (2, –3).

As always, find the y-intercept by putting x = 0. This is at y = –7.

And the line of symmetry is x = 2.

vertex (2, –3)

line of symmetryx = 2

y-intercepty = –7

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y-intercept: y = 02 – 12(0) + 20 So y-intercept is at (0, 20)

x-intercepts: x2 – 12x + 20 = 0Solving for x gives intercepts at (2, 0) and (10, 0)

Vertex and symmetry:x2 – 12x + 20 = (x – 6)2 – 36 + 20= (x – 6)2 – 16So vertex is at (6, –16) andline of symmetry is x = 6

Topic7.4.3

Guided Practice

Solution follows…


1. Sketch the graph of the function y = x2 – 12x + 20, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry.

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Topic7.4.3

Guided Practice

Solution follows…


2. Sketch the graph of the function y = x2 + 8x + 12, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry.

y-intercept: y = 02 + 8(0) + 12 So y-intercept is at (0, 12)

x-intercepts: x2 + 8x + 12 = 0Solving for x gives intercepts at (–6, 0) and (–2, 0)

Vertex and symmetry:x2 + 8x + 12 = (x + 4)2 – 16 + 12= (x + 4)2 – 4So vertex is at (–4, –4) andline of symmetry is x = –4

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Topic7.4.3

Guided Practice

Solution follows…


3. Sketch the graph of the function y = x2 – 2x – 3, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry.

y-intercept: y = 02 – 2(0) – 3 So y-intercept is at (0, –3)

x-intercepts: x2 – 2x – 3 = 0Solving for x gives intercepts at (3, 0) and (–1, 0)

Vertex and symmetry:x2 – 2x – 3 = (x – 1)2 – 1 – 3= (x – 1)2 – 4So vertex is at (1, –4) andline of symmetry is x = 1

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Topic7.4.3

Guided Practice

Solution follows…


4. Sketch the graph of the function y = x2 – 4x – 5, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry.

y-intercept: y = 02 – 4(0) – 5 So y-intercept is at (0, –5)

x-intercepts: x2 – 4x – 5 = 0Solving for x gives intercepts at (5, 0) and (–1, 0)

Vertex and symmetry:x2 – 4x – 5 = (x – 2)2 – 4 – 5= (x – 2)2 – 9So vertex is at (2, –9) andline of symmetry is x = 2

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Topic7.4.3

Guided Practice

Solution follows…


5. Sketch the graph of the function y = –x2 – 2x + 3, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry.

y-intercept: y = –02 – 2(0) + 3 So y-intercept is at (0, 3)

x-intercepts: –x2 – 2x + 3 = 0Solving for x gives intercepts at (–3, 0) and (1, 0)

Vertex and symmetry:–x2 – 2x + 3 = –[(x + 1)2 – 1 – 3]= –(x + 1)2 + 4So vertex is at (–1, 4) andline of symmetry is x = –1

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Topic7.4.3

Guided Practice

Solution follows…


6. Sketch the graph of the function y = –x2 – x + 6, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry.

y-intercept: y = –02 – 0 + 6 So y-intercept is at (0, 6)

x-intercepts: –x2 – x + 6 = 0Solving for x gives intercepts at(–3, 0) and (2, 0)

Vertex and symmetry:

–x2 – x + 6 = – x + – – 6

= – x + +

So vertex is at and line of symmetry is x =

14

12

2 254

12

2

12

– , 254

12

–

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Topic7.4.3

Guided Practice

Solution follows…


7. Sketch the graph of the function y = –2x2 – 8x + 10, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry.

y-intercept: y = –2(0)2 – 8(0) + 10 So y-intercept is at (0, 10)

x-intercepts: –2x2 – 8x + 10 = 0Solving for x gives intercepts at (1, 0) and (–5, 0)

Vertex and symmetry:–2x2 – 8x + 10 = –2[x2 + 4x – 5] = –2[(x + 2)2 – 4 – 5] = –2(x + 2)2 + 18So vertex is at (–2, 18) andline of symmetry is x = –2

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Topic7.4.3

Guided Practice

Solution follows…


8. Sketch the graph of the function y = 2x2 + x – 6, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry.y-intercept: y = 2(0)2 + 0 – 6 So y-intercept is at (0, –6)

x-intercepts: 2x2 + x – 6 = 0Solving for x gives intercepts at

, 0 and (–2, 0)

Vertex and symmetry:

2x2 + x – 6 = 2(x2 + x – 3)

= 2 x + – – 3 = 2 x + –

So vertex is at and line of symmetry is x =

14

2 116

14

– , – 498

14

–

32

12

14

2 498

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Topic7.4.3

Guided Practice

Solution follows…


9. Sketch the graph of the function y = x2 – 4x + 12, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry.y-intercept: y = 02 – 4(0) + 12 So y-intercept is at (0, 12)

x-intercepts: x2 – 4x + 12 = 0This equation doesn’t factor and cannot be solved using the quadratic formula.So there are no x-intercepts.

Vertex and symmetry:x2 – 4x + 12 = (x – 2)2 – 4 + 12= (x – 2)2 + 8So vertex is at (2, 8) andline of symmetry is x = 2

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Topic7.4.3

Guided Practice

Solution follows…


10. Sketch the graph of the function y = 3x2 + 6x + 6, stating the y-intercept and x-intercepts (if appropriate). Use the method of completing the square to find the coordinates of the vertex, and the line of symmetry.y-intercept: y = 3(0)2 + 6(0) + 6 So y-intercept is at (0, 6)

x-intercepts: 3x2 + 6x + 6 = 0This equation doesn’t factor and cannot be solved using the quadratic formula.So there are no x-intercepts.

Vertex and symmetry:3x2 + 6x + 6 = 3(x2 + 2x + 2) = 3{(x + 1)2 – 1 + 2} = 3(x + 1)2 + 3So vertex is at (–1, 3) andline of symmetry is x = –1

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Topic7.4.3

Independent Practice

Solution follows…


In Exercises 1–2, use the information that a ball was

thrown vertically into the air from a platform m above

sea level. The relationship between the height in meters

above sea level, h, and the number of seconds since the

ball was thrown, t, was found to be h = –5t2 + 6t + .

3 2

3 2

1. After how many seconds did the ball reach its maximum height?

2. What was the ball’s maximum height above sea level?

t = s3

5

h = m 33

10

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Topic7.4.3


Solution follows…


3. In the first 8 seconds of its flight, when did the airplane reach its minimum height?

4. What was the minimum height of the plane in the first 8 seconds of its flight?

1 8

The first 8 seconds in the flight of a paper airplane can

be modeled by the quadratic h = t2 – t + 4, where h is

the height in feet and t is the time in seconds.

Use this information to answer Exercises 3–4.

t = 4 s

h = 2 ft

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Topic7.4.3


Solution follows…


5. After how many seconds did the ball reach its maximum height?

6. What was the maximum height of the ball?

7. At what height was the ball initially thrown?

8. When did the ball hit the ground?

A ball is thrown vertically into the air from a platform. The relationship between the ball’s height in meters, h, and the number of seconds, t, since the ball was thrown was found to be h = –5t2 + 10t + 15. Use this information to answer Exercises 5–8.

t = 1 s

h = 20 m

h = 15 m

t = 3 s

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Topic7.4.3

Round UpRound Up


Take a look at Section 7.2 if all this stuff about completing the square seems unfamiliar.

Completing the square is a really useful way of graphing quadratics because it gives you the vertex of the graph straightaway.

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1 Topic 7.4.3 Quadratic Graphs and Completing the Square.

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