Quadratic Functions
Definition of Quadratic Function
A quadratic function is any function specified by a rule that can be written as $ f: x ax^{2} + bx + c $ , where $ a $ . The form $ ax^{2} + bx + c $ is called standard form.
EXAMPLE 12.1 $ f(x)=x^{2} $ , $ f(x)=3x^{2}-2x+15 $ , $ f(x)=-3x^{2}+5 $ , and $ f(x)=-2(x+5)^{2} $ are examples of quadratic functions. $ f(x)=3x+5 $ and $ f(x)=x^{3} $ are examples of nonquadratic functions.
Basic Quadratic Functions
The basic quadratic functions are the functions $ f(x) = x^{2} $ and $ f(x) = -x^{2} $ . The graph of each is a parabola with vertex at the origin (0,0) and axis of symmetry the y-axis (Figs. 12-1 and 12-2).
Graph of a General Quadratic Function
Any quadratic function can be written in the form $ f(x) = a(x - h)^{2} + k $ by completing the square. Therefore, any quadratic function has a graph that can be regarded as the result of performing simple transformations on the graph of one of the two basic functions, $ f(x) = x^{2} $ and $ f(x) = -x^{2} $ . Thus the graph of any quadratic function is a parabola.
EXAMPLE 12.2 The quadratic function $ f(x) = 2x^{2} - 12x + 4 $ can be rewritten as follows:
\[ \begin{aligned}f(x)&=2x^{2}-12x+4\\&=2\left(x^{2}-6x\right)+4\\&=2\left(x^{2}-6x+9\right)-18+4\\&=2\left(x-3\right)^{2}-14\end{aligned} \]
Parabola Opening Up
The graph of the function $ f(x) = a(x - h)^{2} + k $ , for positive a, is the same as the graph of the basic quadratic function $ f(x) = x^{2} $ stretched by a factor of a (if a > 1) or compressed by a factor of 1/a (if 0 < a < 1), and shifted left, right, up, or down so that the point (0,0) becomes the vertex (h,k) of the new graph. The graph of $ f(x) = a(x - h)^{2} + k $ is symmetric with respect to the line x = h. The graph is referred to as a parabola opening up.
Parabola Opening Down
The graph of the function $ f(x) = a(x - h)^{2} + k $ , for negative a, is the same as the graph of the basic quadratic function $ f(x) = -x^{2} $ stretched by a factor of $ |a| $ (if $ |a| > 1 $ ) or compressed by a factor of $ 1/|a| $ (if $ 0 < |a| < 1 $ ), and shifted left, right, up, or down so that the point $ (0,0) $ becomes the vertex $ (h,k) $ of the new graph. The graph of $ f(x) = a(x - h)^{2} + k $ is symmetric with respect to the line x = h. The graph is referred to as a parabola opening down.
Maximum and Minimum Values
For positive a, the quadratic function $ f(x) = a(x - h)^{2} + k $ has a minimum value of k. This value is attained when x = h. For negative a, the quadratic function $ f(x) = a(x - h)^{2} + k $ has a maximum value of k. This value, also, is attained when x = h.
EXAMPLE 12.3 Consider the function $ f(x) = x^{2} + 4x - 7 $ . By completing the square, this can be written as $ f(x) = x^{2} + 4x + 4 - 4 - 7 = (x + 2)^{2} - 11 $ . Thus the graph of the function is the same as the graph of $ f(x) = x^{2} $ shifted left 2 units and down 11 units; see Fig. 12-3.
The graph is a parabola with vertex $ (-2,-11) $ , opening up. The function has a minimum value of -11. This minimum value is attained when x = -2.
EXAMPLE 12.4 Consider the function $ f(x) = 6x - x^{2} $ . By completing the square, this can be written as $ f(x) = -x^{2} + 6x = -(x^{2} - 6x) = -(x^{2} - 6x + 9) + 9 = -(x - 3)^{2} + 9 $ . Thus the graph of the function is the same as the graph of $ f(x) = -x^{2} $ shifted right 3 units and up 9 units. The graph is shown in Fig. 12-4.
The graph is a parabola with vertex $ (3,9) $ , opening down. The function has a maximum value of 9. This value is attained when x = 3.
Domain and Range
The domain of any quadratic function is R, since $ ax^{2} + bx + c $ or $ a(x - h)^{2} + k $ is always defined for any real number x. For positive a, since the quadratic function has a minimum value of k, the range is $ [k, ) $ . For negative a, since the quadratic function has a maximum value of k, the range is $ (- , k] $ .
SOLVED PROBLEMS
12.1. Show that the vertex of the parabola $ y = ax^{2} + bx + c $ is located at $ (-, ) $ .
Completing the square on $ y = ax^{2} + bx + c $ gives, in turn,
\[ \begin{aligned}y&=a\Big(x^{2}+\frac{b}{a}x\Big)+c\\&=a\Big(x^{2}+\frac{b}{a}x+\frac{b^{2}}{4a^{2}}\Big)-\frac{b^{2}}{4a}+c\\&=a\Big(x+\frac{b}{2a}\Big)^{2}+\frac{4ac-b^{2}}{4a}\\ \end{aligned} \]
Thus, the parabola $ y = ax^{2} + bx + c $ is obtained from the parabola $ y = ax^{2} $ by shifting an amount -b/2a with respect to the x-axis and an amount $ (4ac - b^{2})/(4a) $ with respect to the y-axis. Since the vertex of $ y = ax^{2} $ is at (0,0), the vertex of $ y = ax^{2} + bx + c $ is as specified.
12.2. Analyze the intercepts of the graph of $ y = ax^{2} + bx + c $
For x = 0, y = c. Hence, the graph always has one y-intercept, at (0,c).
For y = 0, the equation becomes $ 0 = ax^{2} + bx + c $ . The number of solutions of this equation depends on the value of the discriminant $ b^{2} - 4ac $ (Chapter 5). Thus if $ b^{2} - 4ac $ is negative, the equation has no solutions and the graph has no x-intercepts. If $ b^{2} - 4ac $ is zero, the equation has one solution, x = -b/2a, and the graph has one x-intercept.
If $ b^{2}-4ac $ is positive, the equation has two solutions, $ x= $ and $ x= $ , and the graph has two x-intercepts.
Note that the x-intercepts are symmetrically placed with respect to the line x = -b/2a.
12.3. Show that, for positive a, the quadratic function $ f(x) = a(x - h)^{2} + k $ has a minimum value of k, attained at x = h. For all real x, $ x^{2} $ . Thus, the minimum value of $ (x - h)^{2} $ is 0, and this minimum value is attained at x = h. For positive a and arbitrary k, it follows that:
\[ \begin{aligned}(x-h)^{2}&\geq0\\a(x-h)^{2}&\geq0\\a(x-h)^{2}+k&\geq k\end{aligned} \]
Thus, the minimum value of $ a(x-h)^{2} + k $ is k, attained at x = h.
12.4. Analyze and graph the quadratic function $ f(x) = 3x^{2} - 5 $ .
The graph is a parabola with vertex $ (0,-5) $ , opening up. The graph is the same as the graph of the basic parabola $ y = x^{2} $ stretched by a factor of 3 with respect to the y-axis, and shifted down 5 units. The graph is shown in Fig. 12-5.
12.5. Analyze and graph the quadratic function $ f(x) = -1 - x^{2} $
The function can be rewritten as $ f(x) = -x^{2} - 1 $ . The graph is a parabola with vertex $ (0, -1) $ , opening down. The graph is the same as the graph of the basic parabola $ y = -x^{2} $ compressed by a factor of 3 with respect to the y-axis, and shifted down 1 unit. The graph is shown in Fig. 12-6.
12.6. Analyze and graph the quadratic function \(f(x)=2x^{2}-6x\).
Completing the square, this can be rewritten:
\[ f(x)=2(x^{2}-3x)=2\Big(x^{2}-3x+\frac{9}{4}\Big)-2\cdot\frac{9}{4}=2\Big(x-\frac{3}{2}\Big)^{2}-\frac{9}{2} \]
Hence the graph is a parabola with vertex $ (, -) $ . Opening up. The parabola is the same as the graph of the basic parabola $ y = x^{2} $ stretched by a factor of 2 with respect to the y-axis, and shifted right $ $ units and down $ $ units. The graph is shown in Fig. 12-7.
12.7. Analyze and graph the quadratic function $ f(x)=x^{2}+2x+3 $ .
Completing the square, this can be rewritten:
\[ f(x)=\frac{1}{2}(x^{2}+4x)+3=\frac{1}{2}(x^{2}+4x+4)-2+3=\frac{1}{2}(x+2)^{2}+1 \]
Hence the graph is a parabola with vertex $ (-2,1) $ , opening up. The parabola is the same as the graph of the basic parabola $ y = x^{2} $ compressed by a factor of 2 with respect to the y-axis, and shifted left 2 units and up 1 unit. The graph is shown in Fig. 12-8.
12.8. Analyze and graph the quadratic function $ f(x) = -2x^{2} + 4x + 5 $ .
Completing the square, this can be rewritten as $ f(x) = -2(x - 1)^{2} + 7 $ . Hence the graph is a parabola with vertex $ (1,7) $ , opening down. The parabola is the same as the graph of the basic parabola $ y = -x^{2} $ stretched by a factor of 2 with respect to the y-axis and shifted right 1 unit and up 7 units. The graph is shown in Fig. 12-9.
12.9. State the domain and range for each quadratic function in Problems 12.4–12.8.
In Problem 12.4, the function $ f(x) = 3x^{2} - 5 $ has a minimum value of -5. Therefore, the domain is R and the range is $ [-5, ) $ .
In Problem 12.5, the function $ f(x) = -1 - x^{2} $ has a maximum value of -1. Therefore, the domain is R and the range is $ (-, -1] $ .
In Problem 12.6, the function $ f(x) = 2x^{2} - 6x $ has a minimum value of $ - $ . Therefore, the domain is R and the range is $ $ .
12.10. A field is to be marked off in the shape of a rectangle, with one side formed by a straight river. If 100 feet is available for fencing, find the dimensions of the rectangle of maximum possible area. (See Problem 9.8.)
Let x = length of one of the two equal sides (Fig. 12-10).
In Problem 9.8, it was shown that the area $ A = x(100 - 2x) $ . Rewriting this in standard form, this becomes $ A = -2x^{2} + 100x $ . Completing the square gives $ A = -2(x - 25)^{2} + 1250 $ . Thus the maximum area of 1250 square feet is attained when x = 25. Thus the dimensions are 25 feet by 50 feet for maximum area.
12.11. In the previous problem, what is the domain of the area function $ A(x) $ ? Graph the function on this domain.
The domain of an abstract quadratic function is R, since $ ax^{2} + bx + c $ is defined and real for all real x. In a practical application, this domain may be restricted by physical considerations. Here the area must be positive; hence both x and 100 - 2x must be positive. Thus $ {x R < x < 50} $ is the domain of $ A(x) $ . The graph of $ A = -2(x - 25)^{2} + 1250 $ is the same as the graph of the basic parabola $ y = -x^{2} $ stretched by a factor of 2 with respect to the y-axis and shifted right 25 units and up to 1250 units. The graph is shown in Fig. 12-11.
12.12. A projectile is thrown up from the ground with an initial velocity of $ 144 , ft/sec^{2} $ . Its height $ h(t) $ at time t is given by $ h(t) = -16t^{2} + 144t $ . Find its maximum height and the time when the projectile hits the ground.
The quadratic function $ h(t) = -16t^{2} + 144t $ can be written as $ h(t) = -16 ( t - )^{2} + 324 $ by completing the square. Thus the function attains a maximum value of 324 (when $ t = $ ), that is, the maximum height of the projectile is 324 feet.
The projectile hits the ground when the function value is 0. Solving $ -16t^{2} + 144t = 0 $ or $ -16t(t - 9) = 0 $ yields t = 0 (the starting time) or t = 9. Thus the projectile hits the ground after 9 seconds.
12.13. A suspension bridge is built with its cable hanging between two vertical towers in the form of a parabola. The towers are 400 feet apart and rise 100 feet above the horizontal roadway, while the center point of the cable is 10 feet above the roadway. Introduce a coordinate system as shown.
Find the equation of the parabola in the given coordinate system.
Find the height above the roadway of a point 50 feet from the center of the span.
Since the vertex of the parabola is at $ (0,10) $ , the equation of the parabola can be written as $ y = ax^{2} + 10 $ . At the right-hand tower, 200 feet from the center, the cable is 100 feet high; thus, the point $ (200,100) $ is on the parabola. Substituting yields $ 100 = a(200)^{2} + 10 $ ; hence, a = 90/40,000 or 9/4000. The equation of the parabola is
\[ y=\frac{9x^{2}}{4000}+10 \]
- Here the x-coordinate of the point is given as 50. Substituting in the equation yields
\[ y=\frac{9(50)^{2}}{4000}+\ 10=15.625feet \]
12.14. Find two real numbers whose sum is S and whose product is a maximum.
Let one number be x; then the other number must be S - x. Then the product is a quadratic function of x: $ P(x) = x(S - x) = -x^{2} + Sx $ . By completing the square, this function can be written as $ P(x) = -(x - S/2)^{2} + S^{2}/4 $ . Thus the maximum value of the function occurs when x = S/2. The two numbers are both S/2.
12.15. A salesperson finds that if he visits 20 stores per week, average sales are 30 units per store each week; however, for each additional store that he visits per week, sales decrease by 1 unit. How many stores should he visit each week to maximize overall sales?
Let x represent the number of additional stores. Then the number of visits is given by $ 20 + x $ and the corresponding sales are 30 - x per store. Total sales are then given by $ S(x) = (30 - x)(20 + x) = 600 + 10x - x^{2} $ . This is a quadratic function. Completing the square gives $ S(x) = -(x - 5)^{2} + 625 $ . This has a maximum value when x = 5; thus, the salesperson should visit 5 additional stores, a total of 25 stores, to maximize overall sales.
SUPPLEMENTARY PROBLEMS
12.16. Show that, for negative a, the quadratic function $ f(x) = a(x - h)^{2} + k $ has a maximum value of k, attained at x = h.
12.17. Find the maximum or minimum value and graph the quadratic function $ f(x) = x^{2} + 6x + 9 $
Ans. Minimum value: 0 when x = -3. (See Fig. 12-13.)
12.18. Find the maximum or minimum value and graph the quadratic function $ f(x) = 6x^{2} - 15x $ .
Ans. Minimum value: $ - $ when $ x = $ . (See Fig. 12-14.)
12.19. Find the maximum or minimum value and graph the quadratic function $ f(x) = -x^{2} - x + 6 $ .
Ans. Maximum value: $ $ when $ x = - $ . (See Fig. 12-15.)
12.20. State the domain and range for each quadratic function:
Ans. (a) domain: R, range $ [5,) $ ; (b) domain: R, range $ (-,-7] $ ;
- domain: R, range $ (-,6] $ ; (d) domain: R, range $ [-16,) $
12.21. A projectile is thrown up from an initial height of 72 feet with an initial velocity of 160 ft/sec. Its height $ h(t) $ at time t is given by $ h(t) = -16t^{2} + 160t + 72 $ . Find its maximum height, the time when this maximum height is reached, and the time when the projectile hits the ground.
Ans. Maximum height: 472 feet. Time of maximum height: 5 seconds.
Projectile hits ground: $ 5 + $ seconds.
12.22. 1500 feet of chain link fence are to be used to construct six animal cages as in Fig. 12-16.
Express the total enclosed area as a function of the width x. Find the maximum value of this area and the dimensions that yield this area.
Ans. Area: $ A(x) = x(1500 - 3x) $ . Maximum value: 46,875 square feet. Dimensions: 250 feet by 187.5 feet.
12.23. Find two real numbers whose difference is S and whose product is a minimum.
Ans. S/2 and -S/2
12.24. A basketball team finds that if it charges $25 per ticket, the average attendance per game is 400. For each $.50 decrease in the price per ticket, attendance increases by 10. What ticket price yields the maximum revenue?
Ans. $22.50