3.1 introduction to the family of quadratic functions functions modeling change: a preparation for...

3.1

INTRODUCTION TO THE FAMILY OF QUADRATIC

FUNCTIONS

Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally

Finding the Zeros of a Quadratic Function

Examples 1 and 2 and moreFind the zeros of f(x) = x2 − x − 6.Solution by Factoring f(x) = x2 − x − 6 = (x – 3)(x + 2) = 0 has solutions x = 3 and x = − 2Solution Using the Quadratic Formula

Solution by Graphing


2or 32

251

)1(2

)6)(1(4)1()1(

2

4 22

a

acbbx

3 2 1 1 2 3 4x

6

4

2

2

4

6y

Notice that the graph crosses the horizontal axis at x = 3 and x = − 2

Finding the Zeros of a Quadratic FunctionExample 3The figure shows a graph of y = h(x) = − ½ x2 − 2. What happens if we try to use algebra to find the zeros of h?SolutionWe solve the equation

h(x) = − ½ x2 − 2 = 0. So − ½ x2 = 2

x2 = −4 x = ±

Since is not a real number, there are no real solutions, so h has no real zeros. This corresponds to the fact that the graph of h in the figure does not cross the x-axis.


3 2 1 1 2 3x

5

4

3

2

1

y

44

Concavity and Quadratic FunctionsExample 3Let f(x) = x2. Find the average rate of change of f over the intervals of length 2 starting at x = −4 and ending at x = 4. What do these rates tell you about the concavity of the graph of f?SolutionBetween x = −4 and x = −2, we have

average rate of change of f = (f(-2) – f(-4))/((-2) – (-4)) = – 6.Between x = −2 and x = 0, we have

average rate of change of f = (f(0) – f(-2))/(0 – (-2)) = – 2.Between x = 0 and x = 2, we have

average rate of change of f = (f(2) – f(0))/(2– 0) = 2.Between x = 2 and x = 4, we have

average rate of change of f = (f(4) – f(2))/(4 – 2) = 6.Since the rates of change are increasing, the graph is concave up.


Concavity and Quadratic FunctionsExample 3 continuedGraph of f(x) = x2 showing the average rate of change of f over the intervals of length 2 starting at x = −4 and ending at x = 4. Solution


4 2 0 2 4x

yy = f(x) = x2

Slope = -6

Slope = -2Slope = 2

Slope = 6

Since the rates of change are increasing, the graph is concave up.

Finding a Formula From the Zeros and Vertical Intercept

Example 3Find the equation of the parabola in the figure using the factored form.SolutionSince the parabola has x-intercepts at x = 1 and x = 3, its formula can be written as

f(x) = a(x − 1)(x − 3).Substituting x = 0, y = 6 gives

6 = a(3), resulting in a = 2.Thus, the formula is

f(x) = 2(x − 1)(x − 3). Multiplying out gives f(x) = 2x2 − 8x + 6.


1 3x

6

y

y = f(x)

Formulas for Quadratic Functions

The graph of a quadratic function is a parabola.

• The standard form for a quadratic function is

y = ax2 + bx + c, where a, b, c are constants, a ≠ 0.

The parabola opens upward if a > 0 or downward if a < 0, and it intersects the y-axis at c.

• The factored form, when it exists, is

y = a(x − r)(x − s), where a, r, s are constants, a ≠ 0.

The parabola intersects the x-axis at x = r and x = s.


3.2

THE VERTEX OF A PARABOLA


7 6 5 4 3 2 1 1x

4

2

2

4

6

8

10

12

y

The Vertex Form of a Quadratic FunctionExample 1(a) Sketch f(x) = (x + 3)2 − 4, and indicate the vertex.(b) Estimate the coordinates of the vertex from the graph.(c) Explain how the formula for f can be used to obtain the

minimum of f.Solution(a) (b) The vertex of f appears to be

about at the point (−3,−4).


y = f(x)(c) Note that (x + 3)2 is always positive

or zero, so (x + 3)2 takes on its smallest value when x + 3 = 0, that is, at x = −3. At this point (x + 3)2 − 4 takes on its smallest value, f(−3) = (−3 + 3)2 − 4 = 0 − 4 = −4.

The Vertex Form of a Quadratic Function

The vertex form of a quadratic function is

y = a(x − h)2 + k,

where a, h, k are constants, a ≠ 0.

The graph of this quadratic function has vertex (h, k) and axis of symmetry x = h.To convert from vertex form to standard form, we multiply out the squared term. To convert from standard form to vertex form, we complete the square.


2 4 6x

3

1

1

3

y

The Vertex Form of a Quadratic FunctionExample 2 (a)Put the quadratic function into vertex form by completing the square and then graph it: (a) s(x) = x2 − 6x + 8Solution (a)To complete the square, find the square of half of the coefficient of the x-term, (−6/2)2

= 9. Add and subtract this number after the x-term:

s(x) = x2 − 6x + 8 = x2 − 6x +( 9 – 9) + 8 = (x2 − 6x + 9) – 9 + 8 = (x – 3)2 − 1

The vertex of s is (3,−1) and the axis of symmetry is the vertical line x = 3. The parabola opens upward.


y = s(x)x = 3

(3, -1)

Modeling with Quadratic FunctionsExample 4For t in seconds, the height of a baseball in feet is given by the formula

y = f(t) = −16t2 + 47t + 3.Using algebra, find the maximum height reached by the baseball and the time that height is reached.SolutionThe maximum height is at the vertex, so we complete the square to write the function in vertex form:

y = f(t) = −16(t2 + 47/16 t + 3/16) = −16(t2 + 47/16 t + (-47/32)2 – (-47/32)2 + 3/16) = −16((t – 47/32) 2 − 2401/1024)= −16(t – 47/32) 2 − 2401/64

Thus, the vertex is at the point ( 47/32 , 2401/64 ), or approximately (1.469, 37.516). This means that the ball reaches its maximum height of 37.516 feet at t = 1.469 seconds.


3.1 introduction to the family of quadratic functions functions modeling change: a preparation for...

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