Quadratic Functions

 

Definition of a Quadratic Function

A function f is a quadratic function if f(x) = ax+ bx + c, where a, b, and c are real numbers, with a ¹ 0.

Properties of Graphs of Quadratic Functions

1. The graph of a quadratic function f(x) = ax+ bx + c is called a parabola.

2. If a > 0, the parabola opens upward; if a < 0, the parabola opens downward.

3. As | a| increases, the parabola becomes narrower; as | a| decreases, the parabola becomes wider.

4. The lowest point of a parabola (when a > 0) or the highest point (when a < 0) is called the vertex.

5. The domain of a quadratic function is (-¥ , ¥ ), because the graph extends indefinitely to the right and to the left. If (h, k) is the vertex of the parabola, then the range of the function is [k,¥ ) when a > 0 and (-¥ , k] when a < 0.

6. The graph of a quadratic function is symmetric with respect to a vertical line containing the vertex. This line is called the axis of symmetry. If (h, k) is the vertex of a parabola, then the equation of the axis of symmetry is x = h.

 

Vertex and Intercepts

1. To determine the vertex of the graph of a quadratic function, f(x) = ax+ bx + c, we can either:

a) use the method of completing the square to rewrite the function in the form

2. If a > 0, then the y-coordinate of the vertex represents the minimum value of the function; if a < 0, then the y-coordinate of the vertex represents the maximum value of the function.

3. To find the y-intercept of the graph of f(x) = ax+ bx + c, find f(0); to find the x-intercepts, solve the quadratic equation ax+ bx + c = 0.

Example. For each of the following functions, find the vertex, axis, domain, range, intercepts, and sketch the graph.

a) f(x) = -2(x – 3) + 2

b) f(x) = x - 8x + 10

c) f(x) = 2x+ 12x + 17

Solution. a) According to part 1 of Vertex and Intercepts, a function written in the form

f(x) = a(x – h)+ k has vertex (h, k). So, f(x) = -2(x – 3) + 2 has vertex (3,2).

The axis of symmetry is a vertical line through the vertex, so its equation is x = 3.

The domain is (-¥ , ¥ ). Since a < 0 (a = -2), the range is (-¥ , 2]. (Property 5)

To find the y-intercept, we find f(0).

f(0) = -2(0 – 3)+ 2 = -2(-3)+ 2 = -2(9) + 2 = -18 + 2 = -16

The y-intercept is (0, -16).

To find the x-intercept, we solve the equation, -2(x – 3) + 2 = 0

-2(x – 3) + 2 = 0  
-2(x- 6x + 9) + 2 = 0 Square the binomial
-2x + 12x – 18 + 2 = 0 Simplify
-2x + 12x – 16 = 0  
x - 6x + 8 = 0 Divide both sides by -2
(x – 4)(x – 2) = 0 Solve by factoring

x – 4 = 0 or x – 2 = 0

x = 4 x = 2

  

The x-intercepts are (4,0) and (2,0).

The graph is a parabola. Since a < 0, the parabola opens downward. Since | a| = | -2| = 2, the parabola is narrower than the graph of f(x) = x. (See figure 1, page 276 in text).

To draw the graph, plot the vertex and intercepts; then draw a parabola shape opening downward from the vertex.

b) In this example, we will find the vertex by completing the square.

f(x) = x - 8x + 10  
f(x) = (x - 8x ) + 10 Group terms containing x together
f(x) = (x - 8x + 16 – 16) + 10

Complete the square by adding 16

(-8) = -4; (-4) = 16

You must also subtract 16 so the value of the expression is not changed
f(x) = (x - 8x + 16) – 16 + 10 Regroup
f(x) = (x – 4) - 6 Factor the perfect square

Comparing this result to the form f(x) = (x – h) + k, we see that h = 4 and k = -6; so, the vertex is (4, -6).

The axis of symmetry is x = -4.

The domain is (-¥ , ¥ ). The range is [-6,¥ ).

To find the y-intercept, let x = 0:

f(0) = 0- 8(0) + 10 = 10. So, the y-intercept is (0, 10).

To find the x-intecepts, we will solve the equation x - 8x + 10 = 0

x - 8x + 10 = 0  
x = Quadratic formula
x = Substitute: a = 1, b = -8, c = 10
x = Simplify

x =

x =

x = 4 ±

x » 6.45 or x » 1.55

  

The x-intercepts are (6.45, 0) and (1.55, 0).

The graph is a parabola; since a > 0, the parabola opens upward. Since | a| = 1 > 0, the parabola has the same shape as f(x) = x.

c) We will find the vertex using the formula from part 1 above.

f(x) = 2x + 12x + 17  
x = - Formula for x-coordinate of the vertex
x = - Substitute a = 2 and b = 12

The x-coordinate of the vertex is –3.

To find the y-intercept, we will calculate f(-3).

y = f(-3) = 2(-3) + 12(-3) + 17 = 2(9) + 12(-3) + 17 = 18 – 36 + 17 = -1

So, the vertex is (-3,-1).

The axis of symmetry is x = -3.

The domain is (-¥ ,¥ ). The range is [-1, ¥ ).

To find the y-intercept, let x = 0.

f(0) = 2(0)+ 12(0) + 17 = 17

So, the y-intercept is (0, 17)

To find the x-intercept, solve the equation 2x + 12x + 17 = 0

2x + 12x + 17 = 0  
x = Quadratic formula
x = Substitute a = 2, b = 12, c = 17

x =

x =

x =

x » -2.29 or x » -3.71

 Simplify

The graph is a parabola; since a > 0, the parabola opens upward. Since ½ a½ = ½ 2½ = 2, the parabola is narrower than the graph of f(x) = x.

Applications in which a quantity is to have a maximum or minimum value can often be modeled with quadratic functions. The vertex of the graph is the point of interest because it is at this point that the function has its maximum or minimum value.

Maximum and Minimum

For the function f(x) = ax + bx + c,

1. If a > 0, then f has a minimum that occurs at x = .

2. If a < 0, then f has a maximum that occurs at x = .

Example. The average number of gallons of alcohol consumed by each adult in the United States can be modeled by

f(x) = -0.053x + 1.17x + 35.6

where x represents the number of years after 1970. In what year was alcohol consumption at a maximum? What was the average consumption for that year?

Solution. f(x) = -0.053x + 1.17x + 35.6

a = -0.053 and b = 1.17

Since a < 0, the function has a maximum when x = .

Substituting the values of a and b into this formula, we obtain

x = =

This means that alcohol consumption was at a maximum approximately 11 years after 1970, or in 1981.

To obtain the average consumption for that year, we calculate f(11)

f(11) = -0.053(11) + 1.17(11) = 35.6 » 42.06.

The average consumption for that year was approximately 42 gallons of alcohol by each adult.

 

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