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Foundations of Math 11 Section 7.1 – Quadratic Functions ♦ 301

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7.1 Quadratic Functions

Quadratic functions are found in everyday situations, not just in your math classroom. Tossing a ball in the air is a quadratic function, so is the flow rate of water in a pipe, and the supporting cables of a suspension bridge.

Definition of a Quadratic Function A quadratic function is a function that can be written in the form

f (x) = ax2 + bx + c

where a, b and c are real numbers, and a ≠ c . The graph of a quadratic function is called a parabola.

The simplest quadratic function is the function

y = x2 . This is a basic cup-shaped curve that is

symmetric to the y-axis, which has the equation

x = 0. The lowest point of this cup-shaped curve

(0, 0) is called the vertex. The axis of

symmetry is the line which divides the parabola

into two identical parts.

Parabola

0

2

4

6

8

-3 -2 -1 1 2 3

2xy =

axis of symmetry → x = 0

vertex (0, 0)

y

x

Graphing a Parabola

Step 1: Find the vertex and axis of symmetry.

Step 2: Find the y-intercept by evaluating f (0), and the x-intercept(s) by evaluating f (x) = 0.

Step 3: Add additional points if needed.

Step 4: Sketch graph.

The graph of y = ax2

• The graph of y = ax2 is a parabola with vertex at the origin (0, 0)

• If a > 0: opens upward, vertex is a minimum point.

• If a < 0: opens downward, vertex is a maximum point.

• The parabola will be wide if –1 < a < 1, and narrow if a > 1 or a < –1 (in comparison to the

basic parabola y = x2 ).

• The axis of symmetry is x = 0.

Mt. Douglas Secondary

302 ♦ Chapter 7 – Quadratic Functions Foundations of Math 11


Example 1 Graph of y = ax2

a = 1 a = 3 a = 13

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

a = – 1 a = – 2 a = −12

Solution:▼

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

The graph of y = ax2 + c

• The graph of y = ax2 + c shifts vertically up if c > 0, and vertically down if c < 0.

• The vertex of the parabola is (0, c).

• The axis of symmetry is x = 0.

Example 2 Graph of y = ax2 + c

y = x2 +1 y = x2 −1 y = − 12 x2 +1 Solution:▼

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x




The graph of y = a(x − c)2

• The graph shifts horizontally to the right if c > 0, and horizontally to the left if c < 0.

• The vertex of the parabola is (c, 0).

• The axis of symmetry is x = c.

Example 3 Graph of y = a(x − c)2

y = (x − 2)2 y = (x + 3)2 y = − 12 (x − 2)2 Solution:▼

–6 –4 –2 4 6

6

4

2

0

–2

–4

–6

y

x2

–6 –4 –2 4 62

6

4

2

0

–2

–4

–6

y

x

–6 –4 –2 4 6

6

4

2

0

–2

–4

–6

2

y

x

The graph of y = a(x − h)2 + k

• This quadratic formula is said to be in standard form.

• The vertex of the parabola is (h, k).

• The axis of symmetry is x = h.

• The parabola opens upward if a > 0, and downward if a < 0.

• The parabola will have a minimum value if a > 0, and a maximum value if a < 0.

Example 4 Graph of y = a(x − h)2 + k

y = 2(x −1)2 − 3 y = − 12 (x +1)2 + 2 y = 1

3 (x + 2)2 Solution:▼

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x




Domain and Range

Domain and Range from a Set of Ordered Pairs (x, y) The set of first components, or x-values, in the ordered pairs is the domain. The set of second components, or y-values, is called the range.

Example 1 Determine the domain and range of the ordered pairs:

A = { (1, 2) , (–3, 5) , (4, –2) }

B = { (–3, 4) , (1, 0) , (0, 2) , (3, 2) }

C = { (–2, 1) , (1, 0) , (3, 3) , (1, 4) }

D = { (–3, –1) , (–3, 3) , (2, 3) , (4, 0) }

Solution:▼ Domain of A = { –3, 1, 4 } , Range of A = { –2, 2, 5 }

Domain of B = { –3, 1, 0, 3 } , Range of B = { 0, 2, 4 }

Domain of C = { –2, 1, 3 } , Range of C = { 0, 1, 3, 4 }

Domain of D = { –3, 2, 4 } , Range of D = { –1, 0, 3 }

Example 2 Determine the domain and range of the following parabolas:

a) y = 2(x −1)2 − 3 b) y = − 12 (x +1)2 + 2 c) y = 1

3 (x + 2)2

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

Solution:▼ a) domain: all real numbers

range: y ≥ −3

b) domain: all real numbers

range: y ≤ 2

c) domain: all real numbers

range: y ≥ 0




Example 3

Graph the following quadratic equations, then state the vertex, max/min values of the parabola, axis of symmetry, domain and range. Plot at least 4 points besides the vertex.

a) y = 2(x −1)2 − 3 b) y = − 1

2(x + 2)2 + 4

Solution:▼ a)

y = 2(x −1)2 − 3

= 2(0−1)2 − 3= −1

= 2(2−1)2 − 3= −1

= 2(3−1)2 − 3= 5

= 2(−1−1)2 − 3= 5

x y x y

0 → 0 –1

2 2 –1

3 3 5

–1 –1 5

Vertex: (1, –3)

Max/min: minimum value of –3

Axis of symmetry: x = 1

Domain: all real numbers

Range: y ≥ −3 –6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

b)

y = − 12

(x + 2)2 + 4

= − 12

(−6 + 2)2 + 4 = −4

= − 12

(−4 + 2)2 + 4 = 2

= − 12

(0+ 2)2 + 4 = 2

= − 12

(2 + 2)2 + 4 = −4

x y x y

–6 → –6 –4

–4 –4 2

0 0 2

2 2 –4

Vertex: (–2, 4)

Max/min: maximum value of 4

Axis of symmetry: x = –2


Range: y ≤ 4 –6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

Note: a point horizontally, left or right of the axis of symmetry will always be equal distances.




Finding the Equation of a Parabola from a Graph

Finding the equation of a parabola from a graph requires two things:

1. the vertex;

2. the value that determines the shape and direction of the parabola.

Example 1 Determine an equation for the parabola on the right.

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

Solution:▼ Vertex of the parabola is (–2, 5)

Therefore, y = a(x + 2)2 + 5 .

Now we must solve for the a value which determines the shape of the parabola.

A point on the graph other than the vertex must be found.

Three such points are (0, 3), (–4, 3) or (2, –3). We will use (0, 3).

y = a(x + 2)2 + 5

3= a(0+ 2)2 + 5

3= 4a + 5

4a = −2

a = − 12

Therefore, equation of the parabola is

y = − 1

2(x + 2)2 + 5

Example 2 Determine an equation for the parabola on the right.

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x

Solution:▼ Vertex of the parabola is (1, –4)

Therefore, y = a(x −1)2 − 4 .

Now we must solve for the a value which determines the shape of the parabola.

A point on the graph other than the vertex could be (–2, 2) or (4, 2). We will use (–2, 2).

y = a(x −1)2 − 4

2 = a(–2 −1)2 − 4

2 = 9a − 4

9a = 6

a = 23

Therefore, equation of the parabola is

y = 2

3(x −1)2 − 4




7.1 Exercise Set

Graphing a Parabola

1. Determine whether the graph of each quadratic function opens upwards or downwards.

a) y = 13 x2 + 2

b) y = −2x2 + 3

c) y = −3(x −1)2 + 2

d) y = 2 − 3x2

e) y + 2x − x2 = 0

f) x

2 + 2x + y = 0

2. Graph the quadratic function. Plot at least 4 points other than vertex.

a) y = x2

x y

Vertex: Max/min:

Axis of symmetry: Domain:

Range:

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

b) y = −x2

x y

Vertex: Max/min:


Range:

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x




2. c) y = x2 − 2

x y

Vertex: Max/min:


Range:

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

d) y = (x − 2)2

x y

Vertex: Max/min:


Range:

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

e) y = −(x +1)2 + 2

x y

Vertex: Max/min:


Range:

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

f) y = 12 (x + 2)2 − 2

x y

Vertex: Max/min:


Range:

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x




2. g) y = − 12 (x − 2)2 + 3

x y

Vertex: Max/min:


Range:

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

h) y = −2(x −1)2 − 2

x y

Vertex: Max/min:


Range:

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

i) y − 4 = − 23 (x + 2)2

x y

Vertex: Max/min:


Range:

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

j) y + 3= 34 (x − 4)2

x y

Vertex: Max/min:


Range:

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x




3. Determine an equation for the parabola.

a) 6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

b) 6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

c) 6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

d) 6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

e) 6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

f) 6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x


Foundations of Math 11 Section 7.2 – General Form of a Quadratic Function ♦ 311


7.2 General Form of a Quadratic Function

Not all quadratic functions come in the standard form f (x) = a(x − h)2 + k . Therefore, we must derive a way

of changing a general form quadratic function f (x) = ax2 + bx + c to standard form. Only in standard form can

we easily find the vertex of a quadratic function. The steps are summarized below.

f (x) = ax2 + bx + c

y = ax2 + bx + c

y − c = ax2 + bx

y − c = a(x2 + ba

x + ___ )

Given standard form.

Replace f(x) with y to simplify notation.

Add “–c” to both sides.

Factor “a” out of right-side equation.

y − c + b2

4a= a x2 + b

ax + b2

4a2

⎛

⎝⎜⎞

⎠⎟ Add

b2a

⎛⎝⎜

⎞⎠⎟

2

to right side to make a perfect square,

and a × b

2a⎛⎝⎜

⎞⎠⎟

2

= b2

4a to left side to balance equation.

(Recall perfect squares from Math 10.)

y = a x + b2a

⎛⎝⎜

⎞⎠⎟

2

+ c − b2

4a

f (x) = a x + b2a

⎛⎝⎜

⎞⎠⎟

2

+ c − b2

4a

Factor on right sides and add c − b2

4a to both sides to isolate “y =”.

Write in form f (x) = a(x − h)2 + k .

with h = − b

2a and

k = c − b2

4a

Therefore, the general form quadratic formula f (x) = ax2 + bx + c has vertex

− b

2a, c − b2

4a⎛

⎝⎜⎞

⎠⎟.

We can now define the vertex formula.

The Vertex Formula

The graph of f (x) = ax2 + bx + c, a ≠ 0 is a parabola with vertex (h, k) and axis of

symmetry x = h where h = − b

2a and

k = c − b2

4a or

k = f −b

2a⎛⎝⎜

⎞⎠⎟

.

If a > 0, the parabola has a minimum value and opens upwards. If a < 0, the parabola has a maximum value and opens downwards.




Example 1 Determine the vertex of f (x) = 2x2 − 4x − 3 directly from the equation.

Solution:▼ f (x) = −2x2 + 8x − 3 has a = 2 , b = −4 and c = −3 .

Vertex

− b

2a, c − b2

4a⎛

⎝⎜⎞

⎠⎟= 4

2(2), − 3− (−4)2

4(2)⎛

⎝⎜⎞

⎠⎟= (1, − 5) , or

x-coordinate is − b

2a= 4

2(2)= 1 . So, f (x) = 2x2 − 4x − 3→ f (1) = −2(1)2 − 4(1)− 3= −5 .

Therefore, vertex is (1, -5).

Example 2 Determine the vertex and axis of symmetry for the quadratic function f (x) = 2x2 − 4x −1.

Solution:▼ f (x) = 2x2 − 4x −1 a = 2 , b = −4 and c = −1

Vertex is

− b

2a, c − b2

4a⎛

⎝⎜⎞

⎠⎟= −(−4)

2(2), −1− (−4)2

4(2)⎛

⎝⎜⎞

⎠⎟= (1, − 3) , axis of symmetry is x = 1, or

x-coordinate = − b

2a= −(−4)

2(2)= 1 . So, f (x) = 2x2 − 4x −1→ f (1) = 2(12 )− 4(1)−1= −3 ,

Therefore, vertex is (1, – 3).

Example 3 Graph the quadratic function f (x) = −2x2 + 8x − 3 then state the vertex, axis of symmetry,

max/min, domain and range.

Solution:▼ f (x) = −2x2 + 8x − 3 a = −2 , b = 8 and c = −3

Vertex is

− b

2a, c − b2

4a⎛

⎝⎜⎞

⎠⎟= −8

2(−2), − 3− 82

4(−2)⎛

⎝⎜⎞

⎠⎟= (2, 5) or

x-coordinate = − b

2a= −8

2(−2)= 2 . So, f (x) = −2x2 + 8x − 3→ f (2) = −2(22 )+ 8(2)− 3= 5 ,

Therefore, vertex is (2, 5). Plot at least 4 points other than vertex:

f (x) = −2x2 + 8x − 3

f (0) = −2(02 )+ 8(0)− 3= −3

f (1) = −2(1)2 + 8(1) − 3= 3

f (3) = −2(3)2 + 8(3)− 3= 3

f (4) = −2(4)2 + 8(4)− 3= −3

x f (x)

0 – 3

1 3

3 3

4 – 3

Vertex: (2, 5) Axis of Symmetry: x = 2

Max/min: maximum value of 5


Range y ≤ 5

–6 –4 –2 2 4 6

6

4

2

0

–2

–4

–6

y

x




Example 4 Given that f (x) is a quadratic function with minimum f (1) = −3 , find the vertex, axis of

symmetry, domain and range.

Solution:▼ f (1) = −3 means the point (1, – 3), thus the vertex is (1, – 3).

Axis of Symmetry: x = 1 Domain : all real numbers Range: y ≥ −3

Example 5 Determine a quadratic function with vertex (2, 1) and y-intercept – 3.

Solution:▼ Method 1

The standard form of a quadratic equation is y = a(x − h)2 + k so y = a(x − 2)2 +1 . The y-intercept means where the graph crosses the y-axis, at x = 0; it crosses at (0, – 3).

y = a(x − 2)2 +1

−3= a(0− 2)2 +1

4a = −4

a = −1, thus y = −(x − 2)2 +1

Method 2

The general form of a quadratic equation is y = ax2 + bx + c .

y-intercept –3 has − 3= a(0)2 + b(0)+ c

c = −3, thus y = ax2 + bx − 3

If vertex is (2, 1), then 1= a(2)2 + b(2)− 3

4a + 2b = 4 or 2a + b = 2

By symmetry, if (0, –3) is a point and vertex is (2, 1), then (4, – 3) must be a matching point on the quadratic function. (2 units to the right of x = 2, same y = –3.)

–3= a(4)2 + b(4)− 3

16a + 4b = 0 or b = −4a

2a + b = 2

2a − 4a = 2

a = −1, b = –4a

= –4(−1)

= 4

Thus, y = −x2 + 4x − 3




Example 6 Determine the quadratic function with x-intercepts –1 and 3, and y-intercept 6.

Solution:▼ Method 1

x-intercept is where graph crosses x-axis y = 0, thus (– 1, 0) and (3, 0). y-intercept is where graph crosses y-axis x = 0, thus (0, 6).

y = ax2 + bx + c

6 = a(0)2 + b(0)+ c

c = 6

y = ax2 + bx + c

0 = a(−1)2 + b(−1) = 6 →

0 = a(3)2 + b(3)+ 6

3(a − b = −6) →

9a + 3b = −6

3a − 3b = −189a + 3b = −6

12a = −24a = −2

a − b = −6

−2 − b = −6

b = 4

Thus, quadratic function is y = −2x2 + 4x + 6 .

Method 2

If x-intercepts are –1 and 3, then axis of symmetry is x = −1+ 3

2= 1 and vertex is (1, k).

Thus, y = a(x −1)2 + k with y-intercept of (0, 6)

6 = a(0−1)2 + k

a + k = 6

x-intercept of –1 → (–1, 0) so y = a(x −1)2 + k

0 = a(−1−1)2 + k

k = −4a

Then a + k = 6

a − 4a = 6

a = −2

−2 + k = 6

k = 8

Thus, quadratic function is y = −2(x −1)2 + 8.




7.2 Exercise Set

1. Determine another point that must be on a quadratic function whose graph satisfies the given conditions.

a) Vertex (2, 5); point (4, 1)

b) Vertex (–4, 2), point (3, 5)

c) Vertex (1, –4); point (3, 2)

d) Vertex (–3, 5); point (–6, 2)

e) Vertex (a, b); point (2, 4)

f) Vertex (2, 4); point (a, b)




2. Determine the equation of the axis of symmetry of a quadratic function, given the following information.

a) Vertex (–1, 3)

b) Vertex (4, –1)

c) Points: (2, 5) and (–4, 5)

d) Points: (–3, 2) and (5, 2)

e) Points: (4, 0) and (0, 0)

f) Points: (0, 4) and (–3, 4)

g) Points: (a, b) and (4, b)

h) Points: (a, b) and (c, d)




3. Find the vertex of the following quadratic functions, and state if the vertex is a maximum or minimum point.

a) y = x2 − 2x + 4

b) f (x) = −2x2 − 8x + 3

c) y = 3x2 − 6x +1

d) g (x) = −x2 + 3x − 4

e) h(x) = − 1

3x2 + 2x − 4

f) i (x) = 3x2 − 5x + 2




4. Graph the following quadratic functions. Plot at least 4 points other than the vertex.

a) y = x2 − 4x

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

b) y = 2x2 + 6x

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

c) y = −x2 + 2x + 3

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

d) y = x2 − 6x + 8

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

4. e) y = −x2 + 6x − 9

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

f) y = − 12 x2 + 4x − 6

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x




4. g) y = x2 + 2x + 3

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

h) y = −x2 − 6x − 4

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

i) y = 2x2 − 4x +1

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

j) y = 12 x2 + x − 2

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

k) y = −3x2 −12x − 8

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

l) y = − 13 x2 + 2x − 4

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x




5. Determine a quadratic function with the given information.

a) Vertex (2, 1), and goes through origin

b) Vertex (–2, –5), and y-intercept 3

c) Vertex (–1, 6), and x-intercept – 4

d) Vertex (–4, 0), and f (–2) = 12

e) x-intercepts –1 and 5, minimum value of –1

f) x-intercepts –4 and 2, and y-intercept 4

g) Axis of symmetry is x = 1, y-intercept 2, with only one x-intercept

h) x-intercepts –3 and 0, and f (2) = –6


Foundations of Math 11 Section 7.3 – Quadratic Equations ♦ 333


8. Use the graph to solve x

2 − 3x − 4 = 0.

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

9. Use the graph to solve −x2 + 2x + 3= 0

6

4

2

0

–2

–4

–6

–6 –4 –2 2 4 6

y

x

10. Find the mistake in the working. Then find the correct solution.

a)

(x − 2)(x + 3) = 6

x − 2 = 6 or x + 3= 6

x = 8 or x = 3

b)

x2 = 5x

x = 5


foundations of math 11 section 7.1 – quadratic functions ... · foundations of math 11 section...

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