The result you just proved shows that any function of the form f(x) = ax² + bx + c is the graph of a parabola. Its graph is found by making changes in the graph of the parabola y = x². If a > 1, then the standard parabola is stretched; if a < -1 it is both stretched and reflected in the x-axis. If 0 < a < 1, then the graph of y = x² is contracted; if -1 < a < 0, then the graph of y = x² is reflected in the x-axis and reflected in the x-axis. Thus if a < 1, then we know that the parabola must open downward. The parabola with function f(x) = ax² + bx + c will have its vertex at (-b/(2a), f(-b/(2a))

The vertex for any parabola will have the independent coordinate value of -b/(2a) for its vertex. You can always the find the value for the dependent variable by substituting -b/(2a) for the independent variable.

PROBLEM 1. Graph y = 3x² + 6x + 1 .

The graph of this equation is a parabola that opens upward (because the lead coefficient is 3). The vertex has the x-coordinate x = -6/(2(3)) = -1. The y-coordinate is 3(-1)² +6(-1) + 1 = -2. Thus the parabola opens upward from vertex (-1, - 2). It has the y-intercept of 1 and, by symmetry, must also pass through the point (-2,1)

PROBLEM 2. Graph y = -2x² + 2x - 1

This parabola opens downward. Its vertex has the x-coordinate x = -b/(2a) = - 2/(2(-2)) = 1/2, f(1/2)=-2(1/4) + 2(1/2) - 1 = -1/2. The axis of symmetry is x =1/2

Note that the graph of x = y² is that of a parabola opening to right but also the pair of graphs y = sqrt(x) and y = - sqrt(x). The graph of x = ay² + by + c is a parabola which is the reflection of y = ax² + bx + c across the line y = x so that the parabola "lies on its side". It is stretched away from the y-axis if |a| > 1 or shrunk toward the y-axis if |a| < 1. Its vertex has y = - b/(2a) (with x evaluated by substituting -b/(2a) for y in the equation). The graph opens to the right is a > 0 and to the left if a < 0.

A function f has a maximum value on an interval [a,b] at x = r if and only if f(r) > f(x) for all x in the interval; f has a minimum value on an inteval [a,b] at x = r if and only if f(r) < f(x) for all x in the interval.

PROBLEM 3. A vendor can sell 200 souvenirs per day at a price of $2 each. Each 10¢ price increase decreases the number of sales by 25 per day. Souvenirs cost the vendor $1.50 each. What price should be charged in order to maximize the profit?

x = number of 10¢ increases in price
price = 2 + .10x
# souvenirs sold = 200 - 25x
profit = revenue - cost = (2 + .10x)(200 - 25x) - 1.50(200 - 25x)
= (200 - 25x)(.50 + .10x) = 100 + 7.50x -2.5x²
(domain 0 < x < 8

Parabola opens downward so there is a maximum at x = - 7.5 / (2(-2.5)) = 1.5

Price = $2.15, max profit = $105.625

PROBLEM 4. A trough is to be made by bending a long, flat piece of tin 10 inches wide into a rectangular shape. What depth should the trough be in order to have the maximum possible cross-sectional area?

depth = x inches

width = 10 - 2x inches

A = x(10 - 2x) = 10x - 2x²

max area for x = - 10/(2(-2)) = 2.5 inches (max area = 2.5(5) = 12.5 in²)

Questions and/or comments should be sent to Frances Gulick.
Last modified Sunday, 10-Sep-2000 22:02:06 EDT           © 2009 University of Maryland