Quadratic Functions

 

Functions like  are examples of quadratic functions.  Notice the presence of a quadratic polynomial of the form , A ¹ 0 in each  of the examples of quadratic functions that are presented here.

More generally, a function of the form , where A, B, C are constants, A ¹ 0, is called a quadratic function.  The graph of a quadratic function is called a parabola.  If A > 0, the graph of f opens up and is said to be concave up (this is displayed in Figure 5).  If A < 0, the graph of f opens down and is said to be concave down (this is displayed in Figure 6).  In either case, the indicated point V is called the vertex of the parabola.

                    y                                               y               V                 If A > 0, the vertex is the absolute minimum

                                                                           ·                            of the function, since the graph opens up and

                                                                                                         no point on the graph of f is below the vertex.

 

                                               x                                                 x    If A < 0, the vertex is the absolute maximum

                        ·                                                                              of the function, since the graph opens down

                                V                                                                     and there no point on the graph of f is above

                                                                                                        the vertex.

                 Figure 5                                    Figure 6

The vertex is a point on the xy-plane, so it has the form V = (x, y), where the x-component can be found by using  and the y-component can be found by evaluating .  If the vertex is the absolute maximum of the quadratic function, then the value  is called the maximizer of f and y = is the absolute maximum of f.  If the vertex is the absolute minimum of the quadratic function, then the value  is called the minimizer of f and y = is the absolute minimum of f.      The vertical line with equation  is the axis of symmetry of the parabola (of course, it passes through the vertex).

 

A parabola happens to be one of a group of four curves known as conics.  Conics have various applications and form the trajectory of many objects in the Universe.  The WebSite Occurrence of the Conics (click on it to visit it) by Jill Britton, shows the way conics are obtained by slicing a cone with different planes and also their occurrence and applications in real life.

 

Examples of Analysis of a Quadratic Function:

1)  For the quadratic function ,

 

a)      Explain why the graph of this function is concave down.

Answer:  Since the leading coefficient A = -3 is negative, the graph of f opens down and is concave down.

 

b)      Find the components of the vertex:

Solution:  x component:  ;   y component:    The vertex is the point V = (2, 3).

 

c)      Determine whether the vertex is the absolute maximum or the absolute minimum.

Answer:  Since the graph of f is concave down, the vertex (2, 3) is the absolute maximum of the function.

 

d)      Write the equation of the axis of symmetry.                                             

Solution:  The equation of the axis of symmetry is  x = 2

                                                                                                       f)  Sketch the graph of f and display all the information from parts a)-e). 

e)      Find the intercepts:                                                                                                   y

Solution:  If x = 0, then  y = f(0) = -9  is the y-intercept.                                       4                 (2, 3)

      If y = 0, we must solve the equation    for x.                                           

      Factor completely on the left side to obtain:  -3(x - 1)(x - 3) = 0.                                                             x

      So, x = 1,  x = 3  are the x-intercepts.                                                                                1    2    3  

      The complete list of intercepts includes (0, -9), (1, 0), and (3, 0).

                                                                                                                                                                                                                                                                     - 4                  

                                                                                                                                                                                                                                                                                                              x=2

 

2)  For the quadratic function ,

 

a)      Explain why the graph of this function is concave up.

Answer:  Since the leading coefficient A = 2 is positive, the graph of f opens up and is concave up.

 

b)      Find the components of the vertex:

Solution:  x component:  ;   y component:    The vertex is point V = (1, -18).

 

c)      Determine whether the vertex is the absolute maximum or the absolute minimum.

Answer:  Since the graph of f is concave up, the vertex (1, -18) is the absolute minimum of the function.

 

d)      Write the equation of the axis of symmetry.                                             

Solution:  The equation of the axis of symmetry is  x = 1

                                                                                                       f)  Sketch the graph of f and display all the information from parts a)-e). 

e)      Find the intercepts:                                                                                                   y

Solution:  If x = 0, then  y = f(0) = -16  is the y-intercept.                                                 x = 1              

      If y = 0, we must solve the equation    for x.                                            

      Factor completely on the left side to obtain:  2(x - 4)(x + 2) = 0.                                                             x

      So, x = -2,  x = 4  are the x-intercepts.                                                             -2          1             4  

      The complete list of intercepts includes (0, -16), (-2, 0), and (4, 0).

                                                                                                                                                                                                                                                                                          

                                                                                                                                                                                                                                                                          -16                                        

                                                                                                                                                         (1, -18)

 

3)  Suppose that the total cost (in dollars) of manufacturing q units of a product is given by .

a)      Find the minimizer level of production.

Solution:  The function C is concave up since A = 2 > 0.  Therefore, the vertex of the parabola is the absolute minimum and the minimizer level of production is  units of production.

b)      Find the minimum cost.

Solution:  The minimum cost is

 

4)  The demand equation for a particular product is given by , where p is the selling price (in dollars) per unit, 0 £ p £ 1500, and x is the number of units sold.

a)      Find the quantity x that maximizes the total revenue.  That is, find the maximizer of the revenue R.

Solution:  Recall that total revenue is given by R = px.   In this case, . 

Thus the revenue function is

Notice that in this quadratic function, A =  < 0.  So, the function is concave down and therefore, the vertex is the absolute maximum.

Maximizer:   units sold.

b)      Find the maximum revenue.

      Solution:  The maximum revenue is

c)      Find the unit price p that should be charged in order for the total revenue to be maximized.

      Solution:   per unit.

      Check that  R = px = (750)(150) = $112,500  (this is the maximum revenue)

  

5)      Suppose that a farmer purchases 5000 feet of fencing and plans to fence a rectangular field with the 5000 feet of fencing.  If he does not know College Algebra and chooses at random a rectangle with dimensions 1000 feet by 1500 feet, then the enclosed rectangle has area 1,500,000 squared feet.  Suppose a friend advises him to use a rectangle with dimensions 1100 feet by 1400 feet because with these, the total enclosed area is 1,540,000 squared feet.  Further, suppose that he later discovers that a rectangle with dimensions 1200 feet by 1300 feet still uses the entire 5000 feet of fencing and the enclosed area is bigger than the previous two (area = 1,560,000 squared feet).  Clearly, by changing the dimensions of the rectangle, we obtain a rectangle with perimeter equal to 5000 feet, but the area enclosed changes, depending on the choice of the dimensions.  The question is “what dimensions of the rectangle use the whole 5000 feet of fencing and produce the maximum possible enclosed area?  What is the maximum possible enclosed area?  We can answer these questions using quadratic functions.

Solution:  In the figure below, x represents the length of the rectangle and y represents the width of the rectangle.  Thus, 2x + 2y = 5000 feet 

                                                                                    is the perimeter and A = xy  is the area of the rectangle.  Notice that x + y = 2500 feet can be

                                                                                    obtained from the perimeter.  So,  y = -x + 2500, with 0 £ x £ 2500, so that the width does

                        Area = xy               Width = y               not become negative.  Replace y into the area to get .  Therefore, the

                                                                                    area function in terms of the length x is  .   

                        Length = x                                            Notice that in this function, A = -1 < 0 and therefore the function is concave down.  So, the

                                                                                    maximizer of the area A is  feet.  Therefore, y = 1250 feet also.

The maximum possible enclosed area is  squared feet.  The requested dimensions are 1250 feet by 1250 feet and the maximum possible enclosed area is 1,562,500 squared feet.

 

6)  Suppose that a farmer has 5000 feet of fencing and wants to use it for fencing a rectangular field for which one side is the wall of the farmer’s house. In this case the side along the wall does not require fencing.  What dimensions yield maximum rectangular area and what is the maximum enclosed area?

      Solution:        Wall                 First construct a quadratic function to represent the area of the rectangular plot.

                                                     In Figure 6,  let y represent the length of the side parallel to the wall of the house and let x

                   x                    x          represent the length of each of the other two sides.

                                                     Total amount of fencing:  2x + y = 5000 feet,  so  y = 5000 - 2x,  0 £ x £ 2500.

                               

                               y                    Total Area Enclosed:  A = xy = x(5000 - 2x) squared feet.

          Figure 6                             Quadratic Function:  .

        Notice that this function is concave down (why?), so the vertex is the absolute maximum.  The maximizer is   feet.     (Notice that: 0 £ 1250 £ 2500).

         Therefore,  y = 5000 - 2(1250) =  2500  feet.

         The maximum area enclosed is  squared feet.

         The maximum rectangular area that the farmer can enclose with the 5000 feet of fencing is 3,125,000 squared feet.

 

7)   The manager of a 100-unit apartment complex knows from experience that all units will be occupied if the rent is $800 per month.  A market survey suggests that, on the average, one more unit will become vacant for each $10 increase in the monthly rent.  What monthly rent should the manager charge in order to maximize monthly revenue?  Find the maximum monthly revenue.

        

         SOLUTION:

         Monthly rent (in dollars):      p = 800 + 10x, where x = number of $10 increases).

         Number of occupied units:    q = 100 – x.

         Monthly revenue:    dollars.

         Maximizer of R:  .  This is 10  ten-dollar increases.

         Therefore, an increase of $100 will maximize monthly revenue.

         Maximum monthly revenue:  R (10) = -10(100) + 200(10) + 80000 = $ 81,000.

         Also:  Optimal monthly rent is  p = 800 + 10(10) = $ 900  per month.

                    The optimal number of rented units per month is  q = 100 – 10 = 90  units.

                    The maximum monthly revenue is R = pq = (900)(90) = $ 81,000.

 

8)  An object is thrown straight up from the ground with an initial velocity of 160 feet per second.  Its height h (in feet) from the ground at any time t (in

      seconds), t ³ 0, is given by  . 

a)      When will the object reach maximum height from the ground?

Solution:  Notice that since A = -16 < 0, the function is concave down and therefore, the vertex of the parabola is the absolute maximum.

                 Maximizer:   seconds.  The object will reach maximum height after 5 seconds.

b)      Find the maximum height of the object.

      Solution:  The maximum height is  feet above the ground.

 

NOTE:  The quadratic function ,  A ¹ 0, can be transformed into , where A ¹ 0, and (h, k) is the vertex where  and .    We can derive this as follows:

Completing the square:  . 

If we graph ,  , starting with the graph of  and using transformations, we will see that the extremum of the resulting parabola is (h, k), where  and .  Note:  The standard form of the equation of a parabola with vertex (h, k) and vertical axis of symmetry x = h is .

 

Example:

This is the quadratic function of example 1, above:  . 

Use the method of completing the square to find the vertex of the parabola that it represents.

Solution:         

                        Notice that the given function has been transformed to .

                        Clearly the vertex is the point (2, 3), which agrees with example 1, above.

 

Exercises

For each exercise from 1) to 8), do the following: 

a)      Determine whether the graph is concave up or concave down and justify your answer based on the value of the leading coefficient A.

b)      Find the components of the vertex and state the vertex of the parabola.

c)      Determine whether the vertex is the absolute maximum or the absolute minimum of the function.  Justify your answer.

d)      Write the equation of the axis of symmetry.

e)      Find the intercepts (if any).

f)        Sketch the graph and state the range of the function.

1)  ;                2)  ;           3)  ;                4)  ;

5)  ;                      6)  ;              7)         8)

In exercises 9) and 10), use the method of completing the square to find the vertex of the parabola.  Find the intercepts (if any), the equation of the axis of symmetry, and sketch the graph using transformations on the graph of y = x2.

9)  f(x) = -3x2 + 18x - 26                  10)  f(x) = 2x2 - 6x + 3

 

11)  Suppose that the total cost (in dollars) of manufacturing q units of a product is given by .

        a)   Find the minimizer level of production.            b)  Find the minimum cost.

12)   The demand equation for a particular product is given by , where p is the selling price (in dollars) per unit, 0 £ p £ 8000, and x is the number of units sold.

a)      Find the quantity x that maximizes the total revenue.  That is, find the maximizer of the revenue R.

b)      Find the maximum revenue.

c)      Find the unit price p that should be charged in order for the total revenue to be maximized.

13)   Suppose that a farmer purchases 2000 yards of fencing and plans to use it to fence a rectangular field.  What dimensions of the rectangle use the whole 2000 feet of fencing and produce the maximum possible enclosed area?  What is the maximum possible enclosed area?

14)  Suppose that a farmer has 3000 yards of fencing and wants to use it for fencing a rectangular field for which one side is already fenced by a neighbor’s plot, so that side does not require fencing.  What dimensions of the rectangle yield maximum rectangular area and what is the maximum enclosed area?

15)  A distributor of computer monitors has found that when she sells 200 monitors per week when the price is $800 apiece.  She has found also that she sells 5 more monitors per week for every $10 decrease in the price of each monitor.  What price should she charge per monitor in order to maximize weekly revenue?  Find the maximum weekly revenue.

16)   An object is thrown straight up from the ground with an initial velocity of 80 feet per second.  Its height h (in feet) from the ground at any time t (in

      seconds), t ³ 0, is given by  . 

a)      When will the object reach maximum height from the ground?

b)      Find the maximum height of the object.

 

PROJECT

The total cost, C, of obtaining q units of a particular product can be expressed linearly by the function C(q) = cq + f, (in dollars), where c represents the cost per unit and f represents the fixed cost (setup, overhead, etc.).  The total revenue, R, from the sale of q units of the product can be expressed by the function R(q) = pq, (in dollars), where p represents the unit selling price.  The total profit P after obtaining and selling the q units of the product is the difference of the total revenue minus the total cost and is written P(q) = R(q) – C(q).  The break-even points, BEP, are the values of q that make the total revenue equal to the total cost.  Therefore, the BEP are the same values of q for which P(q) = 0, (no gain and no loss).  

 

The owner of a retail store can obtain a particular type of digital cameras from a manufacturer at a cost of $150 apiece.  The retailer has noticed that if she sells the cameras at the price of $350 apiece, consumers buy 200 cameras in one month.  The retailer’s associated fixed cost amounts to $125 in one month.  In an attempt to increase or stimulate sales, the retailer is planning to lower the unit selling price and estimates that for each $5 reduction in the price, 16 more cameras will be sold during one month. 

a)      Explain why the BEP are the same values of q for which P(q) = 0.

b)      Express the monthly total cost of the digital cameras as a function of q.

c)      Find the linear demand function p = mq + b.  (This is analogous to y = mx + b, the slope-intercept form of the equation of a straight line)

d)      Express the monthly total revenue as a function of q.

e)      Express the retailer monthly profit as a function of q.

f)        Find the optimal quantity q that maximizes the retailer monthly profit.

g)      Find the maximum monthly profit.

h)      Find the optimal selling price per camera.

i)        Find the break-even points.

j)        Sketch the revenue, cost, and profit functions in the same rectangular system.  Display the maximum profit and the BEP clearly.

 

Prepared by: Carlos I. Gil