Math 1113 Notes - Quadratic Functions

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1 Math 1113 Notes - Quadratic Functions Philippe B. Laval Kennesaw State University September 7, 000 Abstract This handout is a review of quadratic functions. It includes a review of the following topics: DeÞnitions and Identities Graph Extreme values Sign of a quadratic function 1 Definitions - Identities 1.1 Definitions Definition 1 (Standard form of a quadratic function) The standard form of a quadratic function is y = f (x) =ax + bx + c where a, b, andc are real constants, a 6= 0. Definition (parabola) quadratic function. A parabola is the name given to the graph of a Definition 3 (quadratic equation) the form ax + bx + c =0. A quadratic equation is an equation of Throughout this chapter, unless speciþed otherwise, a, b, and c will refer to the coefficients which appear in the standard form of a quadratic function. Here are some examples of quadratic functions: y =x +5x 3. Here,a =, b =5, c = 3. y x +x =3. Here, a =1, b =, c =3. (Don t forget to write the quadratic function in standard form before Þguring out what the coefficients are). y = x. Here, a =1, b = c =0. 1

2 1. Identities The following identities often appear when working with quadratic functions. (a + b) = a +b + b (a b) = a b + b (a b)(a + b) =a b These identities are easy to prove, simply use the FOIL method. Various Forms of a Quadratic Function: Completing the Square The two forms we use the most for a quadratic function are: 1. standard form: y = ax + bx + c. vertex form: y = a (x h) + k. Thenumbersh and k correspond to the coordinates of the vertex. More precisely, the vertex is: (h, k)..1 From vertex form to standard form This is done by simply expanding. Example 4 Write y =(x 4) +5 in standard form y = (x 4) +5 = x 8x = x 16x +3+5 = x 16x +37. From standard form to vertex form: completing the square Proposition 5 An expression of the form x +x becomes a perfect square when a is added to it. In other words x +x = x +x + a a This is called completing the square. = x +x + a a = (x + a) a

3 Given a quadratic function, to put it in vertex form, follow the following steps: 1. Factor the coefficient of x from the terms in x and x. Don t do anything with the constant.. Complete the square on the terms in x and x as shown in the proposition above. It amounts to adding and subtracting the square of half the coefficient of x. 3. Distribute the coefficient factored in step 1. Example 6 Write y =x +1x +1 in vertex form. y = x +1x +1 = x +6x +1 we factores the coefficient of x = x +6x we added and subtracted ³ = (x +3) 9 +1 µ 6 = (x +3) = (x +3) 17 3 Solving Quadratic Equations, the Quadratic Formula Solving a quadratic equation, that is an equation of the form ax + bx + c =0 is Þnding the values of x for which the statement ax + bx + c =0is true. Definition 7 The solutions of a quadratic equation ax + bx + c =0are called the roots of the quadratic function f (x) =ax + bx + c. 3.1 Factoring One way of solving a quadratic equation is to factor it. Look at the example below: Example 8 Solve x 5x +6=0 First, we note that x 5x +6=(x ) (x 3). Therefore we have: x 5x +6 = 0 (x ) (x 3) = 0 A product is zero if one of its factors is zero, therefore we get: x = 0 x = 3

4 or x 3 = 0 x = 3 So, x 5x +6=0 has two solutions: x =nd x =3. Remark 9 A quadratic equation has either no solutions, one solution, or two solutions. 3. Quadratic Formula In the example we just did above, we solved the quadratic equation by factoring. Factoring is not always easy. When factoring is not an option, we can use the quadratic formula. This formula will always give you the answer. Definition 10 The solutions to the equation ax + bx + c =0are given by the formula x = b ± b 4ac This formula is called the quadratic formula. You will note that this formula gives two solutions, not just one. Definition 11 The term b 4ac which appears in the quadratic formula is called the discriminant of the quadratic equation. It gives us the following important information: if b 4ac < 0, ax + bx + c =0has no real solutions. if b 4ac > 0, ax + bx + c =0has two real solutions. if b 4ac =0, ax + bx + c =0has one real solutions. Example 1 Solve x x 6=0 We begin by computing the discriminant to Þnd out if the equation has solutions or not. Since a =1, b = 1, andc = 6, wehave b 4ac =( 1) 4 (1) ( 6) =1+4 =5 So, we know the equation has two real solutions. Now, we Þnd the solutions. They are given by x = ( 1) ± 5 = 1 ± 5 So, the solutions are x =3or x = 4

5 30 0 y x Figure 1: Some quadratic functions Example 13 Solve x + x +1=0 This time, a = b = c =1.So, b 4ac =1 4(1)(1) =1 4 = 3 Since the discriminant is negative, the equation has no real solutions. Remark 14 If the right side of the equal sign is not 0, before using the quadratic formula, the equation must be changed so that the right side of the equal sign is 0. For example, to solve x 5x +0=5, Þrst we rewrite it as x 5x +15=0, then solve it. 4 Graph Some quadratic functions are shown in Þgure 1: Looking at the various graphs we have, we can draw several conclusions. There seems to be two kind of graphs. Some open up, the others open down. All the graphs seem to have a similar shape, that of a U. Besides the way they open, another important factor in characterizing these graphs seems to be the position of the graphs with respect to both axes. Some graphs intersect the x-axis twice, others once, some never. All graphs intersect the y-axis once. 5

6 It turns out that it is fairly easy to predict how the graph will look if we have its equation. This is what we study next. 4.1 Orientation If a>0 the parabola opens up. If a<0 the parabola opens down. Example 15 Example 16 The graph of y = x 4 opens up. The graph of y +x 5x =3opens down. 4. The y-intercept Remember, to Þnd the y-intercept, we set x =0and solve for y. If we do this in y = ax + bx + c, wehavethaty = c. So, the y-intercept of y = ax + bx + c is the point (0,c). This tells us that when the equation is in standard form, we get the y-intercept with no computations. Example 17 The y-intercept of y = x 4 is (0, 4) Example 18 The y intercept of y +x 5x =3is (0, 3) 4.3 X-Intercepts Remember, to Þnd the x-intercepts, we set y =0and solve for x. To Þnd the x-intercepts of y = ax + bx + c, weneedtosolveax + bx + c =0. Thisisa quadratic equation, its solutions are given by the quadratic formula. Example 19 Find the x-intercepts of y = x x 6. To Þnd them, we need to solve x x 6=0. We already solved this equation above, and found that its solutions were x = and x =3. Therefore, the x-intercepts are (, 0) and (3, 0). Example 0 Find the x-intercepts of y = x + x +1. To Þnd them, we need to solve x + x +1=0. We already solved this equation above, and found that it had no solutions. Therefore, y = x + x +1 has no x- intercepts. We can even tell more about it. Since the coefficient of x is positive, it means that its graph opens up. Hence, the graph is entirely above the x-axis. 4.4 Vertex Definition 1 (vertex) The vertex is the highest point on a parabola if the parabola opens down, it is the lowest point if the parabola opens up. See Figure. The vertex is a point, hence Þnding the vertex means Þnding its x and y- coordinates. 6

7 Figure : Vertex of a quadratic function The x-coordinate of the vertex is x = b. To Þnd the y-coordinate of the vertex, simply plug the x-coordinate in the equation and compute y. If the quadratic function µ is y = f (x) = ax + bx + c, then the coordinates of the vertex are b µ,f b. If the quadratic function is in vertex form, then the coordinates of the vertex are (h, k). Example Find the vertex of y = x + x +1. Here, a = b = c =1. Hence, the x-coordinate of the vertex is x = b = 1 7

8 The y-coordinate of the vertex is then y = µ 1 + = µ 1 +1 = = 3 4 µ The coordinates of the vertex are: 1, 3. In this case, knowing the vertex 4 also provides information about the x-intercepts. Since this parabola opens up, and its vertex (lowest point) is above the x-axis, the entire parabola has to be above the x-axis. Hence, it cannot have x-intercepts. Example 3 Find the vertex of y = x 8x +16. Here, a =1, b = 8, c =16. The x-coordinate of the vertex is x = 8 =4. The y-coordinate of the vertex is y =4 (8) (4) + 16 = 0. So, the vertex is (4, 0). Notethatsincethey-coordinateofthevertexis0,thevertexisonthe x-axis, hence it is also an x-intercept. Example 4 Find the vertex of y =(x 3) +5 This function is in vertex form, with h =3and k =5. The vertex is (3, 5). Remark 5 If a parabola opens up (a >0) anditsvertexisabovethex-axis (its y-coordinate > 0) then the parabola has no x-intercepts. Remark 6 If a parabola opens down (a <0) and its vertex is below the x-axis (its y-coordinate < 0) then the parabola has no x-intercepts. Remark 7 We can combine the previous two remarks into one which states: If a and the y-coordinate of the vertex have the same sign, the corresponding parabola has no x-intercepts. Remark 8 Finally, if the vertex is on the x-axis (its y-coordinate is 0) then there is only one x-intercept, the same as the vertex. The above remarks tell us that we should always Þnd the vertex before the x-intercepts. In some cases, knowing the vertex allows us to draw conclusions about the x-intercept. 5 Maximum, Minimum and Range Consider the quadratic function y = f (x) =ax + bx + c. Remember that the y-coordinate of a point gives us the vertical position of that point. When a 8

9 parabola opens up, its vertex is the lowest point, hence its y-coordinate corresponds to the smallest value y (i.e. the quadratic function) can have. Similarly, if the parabola opens down, the y-coordinate of its vertex corresponds to the largest value y (i.e. the quadratic function) can have. We summarize this by: If a>0, the quadratic µ function f (x) =ax + bx + c has no maximum. Its minimum is f b (they-coordinateofthevertex). Itsrangeis µ y f b. If a<0, the quadratic µ function f (x) =ax + bx + c has no minimum. Its maximum is f b (the y-coordinate of the vertex). Its range is µ y f b. So, questions related to Þnding the maximum or minimum of a quadratic function are solved by Þnding the vertex of the function. Example 9 Find the maximum, the minimum and the range of f (x) = x + 4x +. Here, a = 1, b =4, c =. The x-coordinate of the vertex is x = 4 ( 1) =. The y-coordinate is +4()+=6. Hence the vertex is (, 6). Since the its graph opens down, f (x) has no minimum. Its maximum is f (x) =6.Itsrange is all reals less than or equal to 6. 6 Transformation of Quadratic Functions All the information given above can be proven by transformation of the graph of y = x. Case 30 The quadratic function is in vertex form. This,means that it is of the form y = a (x h) + k. Consider the following transformations: 1. Startwithy = x. Replace x by x h (horizontal shifting) gives y =(x h). So, the new graph is like that of y = x, simply shifted h units horizontally. 3. Divide y by a, (vertical shrinking or stretching, and possibly reßection) gives y a =(x h) or y = a (x h). The new graph is obtained from the previous one by shrinking or stretching vertically, and also performing a reßection across the y-axis if a<0. 9

10 4. Finally, replace y by y k (vertical shifting) gives y k = a (x h) or y = a (x h) + k. The new graph is obtained from the previous one by shifting vertically by k units. In conclusion, the graph of y = a (x h) + k is obtained by Þrst shifting horizontally the graph of y = x, then performing vertical shrinking or stretching, then performing a reßection across the y-axis if a<0, Þnally, performing a vertical shifting. The vertex of y = x is (0, 0). When we shift the graph horizontally by h units, the new vertex becomes (h, 0). Vertical stretching or shrinking will not change the vertex, since it is on the y-axis. Neither will a reßection across the y-axis. Finally, when we shift the graph k units vertically, the new vertex becomes (h, k). Case 31 The quadratic function is not in vertex form. In this case, we put it in vertex form, then proceed as above. Let s assume that the quadratic function is in standard form. When we put it in vertex form, we obtain: y = ax + bx + c = a µx + ba x + c = a µx ba b + x + 4a b 4a + c µ = a x + b b 4a + c µ = a x + b b 4a + 4ac 4a µ = a x + b + b +4ac 4a Weseeimmediatelythatthex-coordinateofthevertexis b.also, from the remark above, we see that the sign of a determines if there is a reßection across the y-axis. If there is no reßection (a >0), then the graph of y = ax + bx + c has the same orientation as the graph of y = x,itopensup. Otherwise,it opens down. Example 3 Sketch the graph of y = x x +3, Þnd its maximum and minimum. We begin by completing the square. y = x x +3 = x x = x x = (x 1) + The coordinates of the vertex are (1, ). The graph opens up, hence the function has a minimum which is y =. Its graph is obtained by shifting the graph of 10

11 x Figure 3: Graph of y = x x +3 y = x 1 unit to the right, then shifting the resulting graph units up. This transformation is enough to tell us that the graph has no x-intercepts. The shape of the graph is shown in Figure 3. 7 Sign of a Quadratic Function Given a quadratic function y = ax +bx+c, we wish to know where y is positive and where y is negative. Theorem 33 For a given quadratic ax + bx + c, the real solutions of ax + bx + c =0break the real line into intervals. In each interval, ax + bx + c is either always positive, or always negative. Thanks to the theorem, to Þnd the sign of ax + bx + c, we just need to Þnd the sign at one value for each interval. Example 34 Solve x +5x +6> 0 First, we solve x +5x +6=0. The solutions are x = 3 or x = These two values determine three intervals: (, 3), ( 3, ), (, ) Pick a value in the Þrst interval, say -4 and plug it into our quadratic, We get, so our quadratic is positive in the Þrst interval. Pick a value in the second interval, say -.5 and plug it into our quadratic, we get -.5, so our quadratic is negative there. 11

12 Finally, pick a value in the third interval, say 0 and plug it into the quadratic, we get 6, so our quadratic is positive there. Now, we know that x +5x +6 is positive in (, 3) (, ) 8 Problems 1. Do the problems in class. In addition, do the problems below. The problems given in class were: (a) 1,3,5,7,9,11,3,7,36,41,43onpages184,185.. Write each quadratic function in standard form (a) y =(x +3) 0 (b) y =(x 1) 1 (c) y = (x +3) 5 3. Write each equation below in vertex form (a) y =x +1x (b) y = x x (c) y == x 1x 3 4. Solve each equation below (a) 4x +8x +4=0 (b) x + x +=0 (c) x +x +4= 5. For each function below, tell whether the graph opens up or down, Þnd the intercepts, the vertex, the minimum, the maximum, the range and sketch the graph. (a) f (x) =4x +8x +4 (b) g (x) =x + x + (c) h (x) = x +x + 6. Sketch the graphs of y = x x +1and y =4. Find the points where they intersect. 7. Let MPG(t) denote the average miles per gallons of US cars as a function of t, where t is the number of years after Assume the following relation: MPG =0.0075t 0.67t

13 (a) Explain why MPG has a minimum value, then Þnd the minimum value MPG has. (b) Which year will MPG be minimum? 8. When a ball is thrown in the air, its altitude h, t seconds later is given by the relation h = 16t +64t +6. h is expressed in feet. (a) (b) (c) Explain why the ball will reach a maximum altitude. Find the maximum altitude reached by the ball, and how long it takes the ball to reach it. How many seconds after the ball is thrown will it reach the ground. 9. Solve the following quadratic inequalities (a) (x 1) (x +3)< 0 (b) (x 1) (x +3) 0 (c) x +3x 10 > 0 (d) x + x +1< Study the sign of the quadratic functions below (a) f (x) =x +3x 10 (b) g (x) =x + x +1 (c) h (x) =(x 1) (x +3) 13

14 9 Problems - Answers 1. Write each quadratic function in standard form (1 & are related) (a) y =(x +3) 0 (b) y =(x 1) 1 (c) y = (x +3) 5. Write each equation below in vertex form (a) y =x +1x (b) y = x x (c) y == x 1x 3 3. Solve each equation below (a) 4x +8x +4=0(Solutionis:x = 1) (b) x + x +=0(Solution is : No real solutions.) (c) x +x +4= (Solution is : x =1 3 ª, x =1+ 3 ª ) 4. For each function below, tell whether the graph opens up or down, Þnd the intercepts, the vertex, the minimum, the maximum, the range and sketch the graph. (a) f (x) =4x +8x +4 (opens up, y intercept: (0, 4), vertexand x-intercept: ( 1, 0), no max. min is 0, range is all reals 0) (b) g (x) = x + x + (opens up, y intercept (0, ), vertex is ( 0.5, 1.75), no x intercepts, no max., min is 1.75, range is all reals 1.75) (c) h (x) = x +x+ ( opens down. y-intercept is: (0, ), vertex is: (1, 3), x-interceptsare: 1+ 3, 0 and 1 3, 0,nomin, max is 3, range is all reals 3) 5. Sketch the graphs of y = x x +1and y =4. Find the points where they intersect. (points of intersection: (-1,4) and (3,4)) 6. Let MPG(t) denote the average miles per gallons of US cars as a function of t, where t is the number of years after Assume the following relation: MPG =0.0075t 0.67t (a) (b) Explain why MPG has a minimum value, then Þnd the minimum value MPG has. (Answer: minimum value of MPG is 1.4) Which year will MPG be minimum? (Answer: MPG minimum in 1957) 14

15 7. When a ball is thrown in the air, its altitude h, t seconds later is given by the relation h = 16t +64t +6. h is expressed in feet. (a) (b) (c) Explain why the ball will reach a maximum altitude. Find the maximum altitude reached by the ball, and how long it takes the ball to reach it. (Answer: max altitude is 70 feet, reachedins) How many seconds after the ball is thrown will it reach the ground. (Answer: the ball reaches the ground 4.09 s after it is thrown). 8. Solve the following quadratic inequalities (a) (x 1) (x +3)< 0 (Solution: ( 3, 1) ) (b) (x 1) (x +3) 0 (Solution: [ 3, 1] ) (c) x +3x 10 > 0 (Solution: (, 5) (, ) ) (d) x + x +1< 0 (Solution: no solutions ) 9. Study the sign of the quadratic functions below (a) f (x) =x +3x 10 (b) g (x) =x + x +1 (c) h (x) =(x 1) (x +3) 15