## 2.1 Quadratic Functions and Models

(This topic is also in Section 2_1 in Applied Calculus and Section 10.1 in Finite Mathematics and Applied Calculus)

For best viewing, adjust the window width to at least the length of the line below.

### Goodies: On-line Technology

Brush up on your Algebra For all of the tutorials on nonlinear functions and models, you should be familiar with the algebra of exponents and radicals.

The relationship between two quantities is often best modeled by a curved line rather than a straight line. The simplest function with a graph that is not straight line is a quadratic function.

A quadratic function of the variable x is a function that can be written in the form

 f(x) = ax2 + bx + c Function form y = ax2 + bx + c Equation form
where a, b, and c are fixed numbers (with a ≠ 0).
Examples
 1. f(x) = 3x2 - 2x + 1 a = 3, b = -2, c = 1 2. g(x) = -x2 a = -1, b = 0, c = 0 3. h(x) = 3x + 1 Not a quadratic function because a = 0 4. k(x) = x2 - 3x a =   b =   c =

Every quadratic function f(x) = ax2 + bx + c has a parabola as its graph. Whether it opens up ("concave up") or opens down ("concave down") depends on the sign of a . Press the "a > 0" and "a < 0" buttons to see the difference in the figure below.

Features of a Parabola

Features of a Parabola

The graph of f(x) = ax2 + bx + c (a ≠ 0) is a parabola with the following features:

Concavity
If a > 0, the parabola is concave up; if a < 0 it is concave down.

Example
The graph of f(x) = -3x2 - 6x - 3 has a = -3 < 0, so the graph is concave down (press the "a < 0" button above to see its shape.

Vertex
The x coordinate of the vertex is -b/2a. The y coordinate is f(-b/2a).

Example
The graph of f(x) = -3x2 - 6x - 3 has vertex with the following coordinates:

 x coordinate = -b2a = -(-6)2(-3) = -1
y coordinate =  f -b2a
= f(-1)  =   -3(-1)2 - 6(-1) - 3   =   0

y-Intercept
The y-intercept is given by y = c.

Example
The graph of f(x) = -3x2 - 6x - 3 has c = -3, so the y-intercept is given by y = -3.

x-Intercepts
The x-intercepts, if they exist, are given by setting

ax2 + bx + c = 0
ans solving for x. To do this, you may beed to factor the quadratic (better if it works!) or use the quadratic formula:

x =
 - b b2 - 4ac

2a
For a quick review, see Solving Polynomial Equations in the on-line algebra review.
Note: If the quadratic has no real factors (equivalently, if b2 - 4ac is negative) then there are no x-intercepts -- the parabola is entirely above or below the x-axis.

Example
The graph of f(x) = -3x2 - 6x - 3 has x-intercept obtained by solving the quadratic

-3x2 - 6x - 3 = 0
To solve, first divide both sides by -3 and then factor:
 -3x2 - 6x - 3 = 0 x2 + 2x + 1 = 0 (x + 1)(x + 1) = 0
The only solution is x = 1, so this gives the only x-intercept.

y = -3x2 - 6x - 3

Let f(x) = 4x2 - 8x - 21.

1. Compute the coordinates of the vertex of its graph and click on its location in the following grid.

2. Now compute the y-intercept.

 y-intercept =

3. Now Compute the x-intercepts, and click on both of them in the grid below. (Try to be as accurate as possible.)

Let f(x) = -x2 + x - 3.

 a =   b =   c =
 x-coordinate of vertex: Enter "none" if there isn't one. y-coordinate of vertex: Enter "none" if there isn't one.
 y-intercept: Enter "none" if there isn't one.
 x-intercept(s): Separate by commas; enter "none" if there aren't any.

Now select (click on) the correct graph from the following (the gridlines are one unit apart):

The population of Roman Catholic nuns in the US during the last 25 years of the 90's can be modeled by
P(t) = -0.15t2 + 2t + 123.75 thousand nuns,       (5 ≦ t ≦ 25)
where t = 0 represents January 1970 (so that January 1975 is represented by t = 5).*

*Figures are rounded. Source: Center for Applied Research in the Apostolate/The New York Times, January 16, 2000, p. A1.

 According to the model, the year during which the population of Roman Catholic nuns in the US reach a peak was .

 According to the model, the largest population of Roman Catholic nuns in the US was (rounded to the neartest 1000).

Now graph the curve -0.15x^2 + 2x + 123.75 on the Function Evaluator & Grapher to confirm your calculations.

You now have several options

• Try some of the questions in the true/false quiz (warning: it covers the whole Chapter) by pressing the button on the sidebar.
• Try some of the on-line review exercises (press the "Review Exercises" button on the side. Again, these questions cover the whole chapter, but you should now be able to answer 1 through 3.)
• Try some of the exercises from Section 2.1 in Applied Calculus or Section or Section 10.1 of Finite Mathematics and Applied Calculus.

Top of Page

Last Updated: March, 2007