The Graph of a Quadratic FunctionA quadratic function is a polynomial function of degree 2 which can be written in the general form, Show f(x)=ax2+bx+ c Here a, b and c represent real numbers where a≠0. The squaring function f(x)=x2 is a quadratic function whose graph follows. This general curved shape is called a parabolaThe U-shaped graph of any quadratic function defined by f(x)=ax2+bx +c, where a, b, and c are real numbers and a≠0. and is shared by the graphs of all quadratic functions. Note that the graph is indeed a function as it passes the vertical line test. Furthermore, the domain of this function consists of the set of all real numbers (−∞,∞) and the range consists of the set of nonnegative numbers [0,∞). When graphing parabolas, we want to include certain special points in the graph. The y-intercept is the point where the graph intersects the y-axis. The x-intercepts are the points where the graph intersects the x-axis. The vertexThe point that defines the minimum or maximum of a parabola. is the point that defines the minimum or maximum of the graph. Lastly, the line of symmetryThe vertical line through the vertex, x=−b2a, about which the parabola is symmetric. (also called the axis of symmetryA term used when referencing the line of symmetry.) is the vertical line through the vertex, about which the parabola is symmetric. For any parabola, we will find the vertex and y-intercept. In addition, if the x-intercepts exist, then we will want to determine those as well. Guessing at the x-values of these special points is not practical; therefore, we will develop techniques that will facilitate finding them. Many of these techniques will be used extensively as we progress in our study of algebra. Given a quadratic function f(x)=ax2+bx+c, find the y-intercept by evaluating the function where x =0. In general, f(0)=a(0)2+b(0)+c=c, and we have y-intercept(0, c) Next, recall that the x-intercepts, if they exist, can be found by setting f(x)=0. Doing this, we have a2+b x+c=0, which has general solutions given by the quadratic formula, x=−b±b2−4ac2a. Therefore, the x-intercepts have this general form: x-intercepts(−b− b2−4ac2a,0)and(−b+b2−4ac 2a,0) Using the fact that a parabola is symmetric, we can determine the vertical line of symmetry using the x-intercepts. To do this, we find the x-value midway between the x-intercepts by taking an average as follows: x=(−b−b2−4ac2a+−b+b2 −4ac2a)÷2=(−b−b2−4ac −b+b2−4ac2a)÷(21)=− 2b2a⋅12=−b2a Therefore, the line of symmetry is the vertical line x =−b2a. We can use the line of symmetry to find the the vertex. Lineof symmetryVertexx=−b2a (−b2a,f(−b2a)) Generally three points determine a parabola. However, in this section we will find five points so that we can get a better approximation of the general shape. The steps for graphing a parabola are outlined in the following example. Example 1Graph: f(x)=−x2− 2x+3. Solution: Step 1: Determine the y-intercept. To do this, set x=0 and find f(0). f(x)=−x2−2x+3f(0)= −(0)2−2(0)+3=3 The y-intercept is (0,3). Step 2: Determine the x-intercepts if any. To do this, set f(x)=0 and solve for x. f(x)=−x2−2x+3Setf(x)=0.0=−x2−2 x+3Multiplybothsidesby−1 .0=x2+2x−3Factor .0=(x+3)(x−1) Seteachfactorequaltozero. x+3=0orx−1=0x=−3x=1 Here where f(x)=0, we obtain two solutions. Hence, there are two x-intercepts, (−3,0) and (1, 0). Step 3: Determine the vertex. One way to do this is to first use x=−b2a to find the x-value of the vertex and then substitute this value in the function to find the corresponding y-value. In this example, a=−1 and b=−2. x= −b2a=−(−2)2(−1) =2−2=−1 Substitute −1 into the original function to find the corresponding y-value. f(x)=−x2−2x+3f(−1)=−(−1)2−2(−1)+3=−1+2+3 =4 The vertex is (−1,4). Step 4: Determine extra points so that we have at least five points to plot. Ensure a good sampling on either side of the line of symmetry. In this example, one other point will suffice. Choose x=−2 and find the corresponding y-value. xyPoint−23f(−2)=−(−2)2−2(−2)+ 3=−4+4+3=3(−2,3) Our fifth point is (−2, 3). Step 5: Plot the points and sketch the graph. To recap, the points that we have found are y-interc ept:(0,3)x-intercepts: (−3,0) and (1,0)Vertex: (−1,4)Extrapoint:(−2, 3) Answer: The parabola opens downward. In general, use the leading coefficient to determine if the parabola opens upward or downward. If the leading coefficient is negative, as in the previous example, then the parabola opens downward. If the leading coefficient is positive, then the parabola opens upward. All quadratic functions of the form f(x)=ax2+bx +c have parabolic graphs with y-intercept (0,c). However, not all parabolas have x-intercepts. Example 2Graph: f (x)=2x2+4x+5. Solution: Because the leading coefficient 2 is positive, we note that the parabola opens upward. Here c = 5 and the y-intercept is (0, 5). To find the x-intercepts, set f (x)=0. f(x)=2x2+4x+50=2x2+4x+5 In this case, a = 2, b = 4, and c = 5. Use the discriminant to determine the number and type of solutions. b2−4ac=(4)2−4(2)(5)=16 −40=−24 Since the discriminant is negative, we conclude that there are no real solutions. Because there are no real solutions, there are no x-intercepts. Next, we determine the x-value of the vertex. x=−b2a=−(4)2 (2)=−44=−1 Given that the x-value of the vertex is −1, substitute −1 into the original equation to find the corresponding y-value. f(x)=2x2+4x+5 f(−1)=2(−1)2+4(−1)+5=2−4+5=3 The vertex is (−1, 3). So far, we have only two points. To determine three more, choose some x-values on either side of the line of symmetry, x = −1. Here we choose x-values −3, −2, and 1. xy Points−311f(−3)=2(−3)2+4(−3)+5=18−12+5=11(−3,11 )−25f(−2)=2(−2 )2+4(−2)+5=8−8+5=5(−2,5)111f(1)=2(1)2+4(1 )+5=2+4+5=11(1,11) To summarize, we have y-intercept:(0,5)x-intercepts:NoneVertex: (−1,3)Extrapoints: (−3,11),(−2,5),(1,11) Plot the points and sketch the graph. Answer: Example 3Graph: f(x)=x2−2x−1. Solution: Since a = 1, the parabola opens upward. Furthermore, c = −1, so the y-intercept is (0,−1). To find the x-intercepts, set f(x)=0 . f(x)=x2−2x−10= x2−2x−1 In this case, solve using the quadratic formula with a = 1, b = −2, and c = −1. x=−b±b2−4ac2a=−(− 2)±(−2)2−4(1)(−1) 2(1)=2±82=2±222=2(1±2)2 =1±2 Here we obtain two real solutions for x, and thus there are two x-intercepts: (1−2,0)and(1+2,0 )Exactvalues(−0.41,0) (2.41,0)Ap proximatevalues Approximating the x-intercepts using a calculator will help us plot the points. However, we will present the exact x-intercepts on the graph. Next, find the vertex. x=−b2a=−(−2) 2(1)=22=1 Given that the x-value of the vertex is 1, substitute into the original equation to find the corresponding y-value. y=x2−2x−1= (1)2−2(1)−1=1−2−1=−2 The vertex is (1, −2). We need one more point. xyPoint2−1 f(2)=(2)2−2(2)−1=4−4−1=−1 (2,−1) To summarize, we have y-intercept:(0,−1)x-in tercepts:(1−2,0)and(1+2,0) Vertex:(1,−2)Extra point:(2,−1) Plot the points and sketch the graph. Answer: Try this! Graph: g(x)=−4x2+12x−9. Answer: Finding the Maximum or MinimumIt is often useful to find the maximum and/or minimum values of functions that model real-life applications. To find these important values given a quadratic function, we use the vertex. If the leading coefficient a is positive, then the parabola opens upward and there will be a minimum y-value. If the leading coefficient a is negative, then the parabola opens downward and there will be a maximum y-value. Example 4Determine the maximum or minimum: y=−4 x2+24x−35. Solution: Since a = −4, we know that the parabola opens downward and there will be a maximum y-value. To find it, first find the x-value of the vertex. x=−b2ax-valueoftheve rtex.=−242(−4) Substitutea=-4andb=24.=−24−8Simplify. =3 The x-value of the vertex is 3. Substitute this value into the original equation to find the corresponding y-value. y=−4x2+24x−35Substitute x=3.=−4(3)2+24(3)−35 Simplify.=−36+72−35 =1 The vertex is (3, 1). Therefore, the maximum y-value is 1, which occurs where x = 3, as illustrated below: Note: The graph is not required to answer this question. Answer: The maximum is 1. Example 5Determine the maximum or minimum: y =4x2−32x+62. Solution: Since a = 4, the parabola opens upward and there is a minimum y-value. Begin by finding the x-value of the vertex. x=−b2a=−− 322(4)Substitutea=4an db=-32.=−−328Sim plify.=4 Substitute x = 4 into the original equation to find the corresponding y-value. y=4x2−32x+62=4( 4)2−32(4)+62=64−128+62=−2 The vertex is (4, −2). Therefore, the minimum y-value of −2 occurs where x = 4, as illustrated below: Answer: The minimum is −2. Example 6The height in feet of a projectile is given by the function h(t)=−16t2+72t, where t represents the time in seconds after launch. What is the maximum height reached by the projectile? Solution: Here a=−16, and the parabola opens downward. Therefore, the y-value of the vertex determines the maximum height. Begin by finding the time at which the vertex occurs. t=−b2a=− 722(−16)=7232=94 The maximum height will occur in 94 seconds (or 214 seconds). Substitute this time into the function to determine the maximum height attained. h(94 )=−16(94)2+72(94) =−16(8116)+72(94)= −81+162=81 Answer: The maximum height of the projectile is 81 feet. Finding the Vertex by Completing the SquareIn this section, we demonstrate an alternate approach for finding the vertex. Any quadratic function f(x)=ax2+bx+c can be rewritten in vertex formA quadratic function written in the form f(x)=a(x−h)2+k., f(x)=a (x−h)2+k In this form, the vertex is (h,k). To see that this is the case, consider graphing f(x)=(x−2)2+3 using the transformations. y= x2Basicsquaringfunctiony=(x−2)2Horizontalshift right2unitsy=(x−2)2+3Verticalshiftup3units Use these translations to sketch the graph, Here we can see that the vertex is (2, 3). f(x)= a(x−h)2+k↓ ↓f(x)=(x−2)2+3 When the equation is in this form, we can read the vertex directly from it. Example 7Determine the vertex: f(x)=2(x+3)2−2. Solution: Rewrite the equation as follows before determining h and k. f(x)=a(x−h )2+k↓↓f(x)=2[x−(−3)]2+(−2) Here h = −3 and k = −2. Answer: The vertex is (−3, −2). Often the equation is not given in vertex form. To obtain this form, complete the square. Example 8Rewrite in vertex form and determine the vertex: f(x)= x2+4x+9. Solution: Begin by making room for the constant term that completes the square. f(x)= x2+4x+9=x2+4x+___+9−___ The idea is to add and subtract the value that completes the square, (b2)2, and then factor. In this case, add and subtract (42)2=(2) 2=4. f(x)=x2+4x+9Ad dandsubtract4.=x2+4x+4+9 −4Factor.=(x2+4x+4)+5. =(x+3)(x+2)+5 =(x+2)2+5 Adding and subtracting the same value within an expression does not change it. Doing so is equivalent to adding 0. Once the equation is in this form, we can easily determine the vertex. f(x)=a(x−h)2+k ↓↓f(x)=(x −(−2))2+5 Here h = −2 and k = 5. Answer: The vertex is (−2, 5). If there is a leading coefficient other than 1, then we must first factor out the leading coefficient from the first two terms of the trinomial. Example 9Rewrite in vertex form and determine the vertex: f(x)=2x2−4x+8. Solution: Since a = 2, factor this out of the first two terms in order to complete the square. Leave room inside the parentheses to add and subtract the value that completes the square. f(x) =2x2−4x+8=2(x2−2x)+8 Now use −2 to determine the value that completes the square. In this case, (−22)2=(−1)2=1. Add and subtract 1 and factor as follows: f(x)=2x2−4x+8=2(x2−2x+__−__)+8Addandsubtract1. =2(x2−2x+1−1)+8Factor .=2[(x−1)(x−1)−1]+8=2[(x−1)2−1]+8Distributet he2.=2(x−1)2−2+8 =2(x−1)2+6 In this form, we can easily determine the vertex. f(x)=a(x−h)2+k ↓↓f(x)=2(x−1)2+ 6 Here h = 1 and k = 6. Answer: The vertex is (1, 6). Try this! Rewrite in vertex form and determine the vertex: f(x)=− 2x2−12x+3. Answer: f(x)=−2(x+3)2+21; vertex: (−3,21) Key Takeaways
Topic Exercises
Part A: The Graph of Quadratic FunctionsDoes the parabola open upward or downward? Explain. Determine the x- and y-intercepts. Find the vertex and the line of symmetry. Graph. Find the vertex and the y-intercept. In addition, find the x-intercepts if they exist.
Part B: Finding the Maximum or MinimumDetermine the maximum or minimum y-value. Given the following quadratic functions, determine the domain and range.
Part C: Finding the Vertex by Completing the SquareDetermine the vertex. Rewrite in vertex form y=a(x−h )2+k and determine the vertex. Graph. Find the vertex and the y-intercept. In addition, find the x-intercepts if they exist.
Part D: Discussion BoardAnswers
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