Lesson 7

Sometimes you will encounter a question in which the quadratic function is not written in standard, factored, or vertex form. In these cases, a graphing calculator can also be a great help. For examples that use a graphing calculator, see “Example 1: Verifying solutions to a quadratic equation” on pages 397 and 398 and “Example 3: Solving a quadratic equation in non-standard form” on page 401 of the textbook. The solution to “Example 1” finds the x-intercepts, whereas the solutions to “Example 3” are the x-coordinates of the points where the two graphs intersect.

 

Example 1: In the Cafeteria

 

A cafeteria owner is planning a new cafeteria in a local hospital. She believes that the monthly profit of a cafeteria can be modelled by the quadratic function P(x) = −0.25x2 + 17.5x + 1500, where x is the seating capacity and P(x) is the profit. How many seats would the cafeteria need in order to earn a monthly profit of $1700 or more?

 

Since you know that the owner is looking for a profit of at least $1700, replace P(x) with 1700.

 

1700 = −0.25x2 + 17.5x + 1500

 

Input the following values into your graphing calculator:

 

Y1 = 1700

 

Y2 = −0.25x2 + 17.5x + 1500

 

 

Change the window settings to show the intersection points. Sometimes this takes a bit of trial and error.

 

 

Now find the points of intersection using the CALCULATE feature (2nd and TRACE); choose option “5: Intersect” from the menu.

 

 

There are two points of intersection: x = 14.38 and x = 55.6.

 

Since the problem is asking about seats in a cafeteria, fractional numbers don’t make sense. So, you would use x = 15 and x = 55 as your solutions.

 

The owner would earn a profit of $1700 starting at 14.38 seats; and she would earn $1700 or more until 55.6 seats were filled.

 

The answer to the question is, yes, she could earn a monthly profit of $1700 or more if between 15 and 55 seats were filled. Note: You would round up for 14.38 because 14 is not enough to get to the line and you would round down for 55.6 because the parabola is below the line after 56.

 

Example 2: Parabolic Reflectors

 

View The Structure of Parabolic Reflectors for an example of using a quadratic function to model an object.

 

 

Self-Check 3

 

Johan made a long pass down court to Yuan. Alise, a friend from Johan’s math class, calculated that the height of the pass was a function of the distance from the thrower.

 

Alise’s function is h(x) = −0.136d2 + 1.63d + 1.67, where d is the distance and h is the height of the ball in metres. The height of Yuan’s hands catching the ball matched the function h = −0.21(d −10.5) + 2.5.

 

At what height and at what distance away from Johan did Yuan catch the ball? Solve by graphing the two functions with a graphing calculator or some other graphing tool. Round the answer to one decimal place. Answer

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If you haven’t done so already, you may want to add what you have learned about quadratic functions in this lesson to your notes organizer.

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If you feel you need a bit more practice, complete Self-Check 4 before moving to Connect.

 

Self-Check 4

Complete all or parts of questions 4, 7, 8, and 13 on pages 411 to 413 and questions 2, 3, 7, 8, and 9 on pages 402 and 403 of the textbook. When you finish a question, check your work using the shortened answers given on pages 560 to 564 of the textbook. If you are still unclear about how to answer any of the questions, ask your teacher for help.