


Quadratic relation, parabolas : translations and applications
PARABOLAS: TRANSLATIONS AND APPLICATIONS QUADRATIC RELATION A quadratic relation in two variables is a relation that can be written in the form y=ax^{2}+bx+c or x=ay^{2}+by+c where a, b, and c are real numbers, and a≠0. Example 1 GRAPHING THE SIMPLEST QUADRATIC RELATION Graphy=x^{2} Notice in Figure 3.16 that the part of the graph in quadrant II is a “mirror image” of the part in quadrant I. We say that this graph is symmetric with respect to the yaxis. (More will be said about symmetry in the next section.) The line of symmetry for a parabola is the axis of the parabola. The lowest point on this parabola, the point (0,0), is called the vertex of the parabola. Starting with y=x^{2}, there are several possible ways to get a more general Example 2 GRAPHING RELATIONS OF THE FORM y=ax^{2} Graph each relation. (a) y=2 x^{2} A table of selected ordered pairs is given with the graph in Figure 3.17. The yvalues of the ordered pairs of this relation are twice as large as the corresponding yvalues for the graph of y=x^{2}. This makes the graph rise more rapidly, so the parabola is narrower than the parabola for y=x^{2}, as can be seen in the figure.3.17
Let’s see how our solver generates graph of this and similar problems. Click on "Solve Similar" button to see more examples. (b) y=− 12 x^{2}
Let’s see how our solver generates graph of this and similar problems. Click on "Solve Similar" button to see more examples. As Example 2 suggests, a in y=ax^{2} determines the width of a parabola, so that it is narrower than the graph of y=x^{2}. when a>1 and broader than the graph of y=x^{2} when a<1. The next two examples show how changing y=x^{2} to y=x^{2}+k or to GRAPHING A RELATION OF THE y=x^{2}+k Graph y=x^{2}4
the yintercept is − 4, which is the yvalue of the vertex. The xintercepts are found by setting y=0: y=x^{2}4 0=x^{2}4 4=x^{2} x=2 or x=− 2 The vertical shift of the graph in Example 3 is called a translation. Example 4 below shows a horizontal translation, which is a shift to the right or left. Example 4 GRAPHING A RELATION OF THE FORM y={xh}^{2} Graph y=(x4)^{2} A combination of all the transformations illustrated in Examples 2, 3, and 4 is shown in the next example. Example 5 GRAPHING A RELATION OF THE FORM y=a (xh)^{2}+k Graph y=− (x+3)^{2}+1. This parabola is translated 3 units to the left and 1 unit up. Because of the negative sign, it opens downward, so that the vertex, the point (− 3,1), is the highest point on the graph. The axis is the line x=− 3. The yintercept is y=− (0+3)^{2}+1=− 8 . By symmetry about the axis x=− 3, the point (− 6,− 8) also is on the graph. The xintercepts are found by solving the equation 0=− (x+3)^{2}+1 0=− (x^{2}+6 x+9)+1 Square the binomial.
Examples 25 suggest the following generalizations. Let’s see how our solver generates graph of this and similar problems. Click on "Solve Similar" button to see more examples. GRAPH OF A PARABOLA The graph of y=a (xh)^{2}+k, where a≠0, Example 6 GRAPHING PARABOLA BY COMPLETING THE SQUARE Graph y=− 3 x^{2}2 x+1. y− 3+13+19=x^{2}+23 x+19 [12 (23)]^{2}=19, so add 1/9 to both side. y− 3=(x+13)^{2}49 Subtract 49. y=− 3 (x+13)^{2}+43 Multiply by − 3. or y=− 3 [x(− 13)]^{2}+43
Figure 3.22 Now the equation of the parabola is written in the form y=a (xh)^{2}+k, and this rewritten equation shows that the axis of the parabola is the vertical line x=− 13 and that the vertex is (− 13,43). Use these results, together with the intercepts and additional ordered pairs as needed, to get the graph in Figure 3.22. From the graph, the domain of the relation is (− ∞,∞) and the range is (− ∞,43). Let’s see how our solver generates graph of this and similar problems. Click on "Solve Similar" button to see more examples. A formula for the vertex of the graph of the quadratic relation y=ax+bx+c can be found by completing the square for the general form of the equation. y=ax^{2}+bx+c a≠0 ya=x^{2}+ba x+ca Divide by a yaca=x^{2}+ba x Subtract ca. yaca+b^{2}4 a^{2}=x^{2}+ba x+b^{2}4 a^{2} Add b^{2}4 a^{2} ya+b^{2}4 ac4 a^{2}=(x+b2 a)^{2} Combine terms on left and factor on right. ya=(x+b2 a)^{2}b^{2}4 ac4 a^{2} Get y term alone on the left. y=a (x+b2 a)^{2}+4 acb^{2}4 a Multiply by a
The ﬁnal equation shows that the vertex (h,k) can be expressed in terms of a,b, and c. However, it is not necessary to memorize the expression for k. since it can be obtained by replacing x with − b2 a. VERTEX OF A PARABOLA Y=ax^{2}+bx+c The xvalue of the vertex of the parabola y=ax^{2}+bx+c, where a≠0, is − b2 a. Example 7 USING THE VERTEX FORMULA Use the formula above to ﬁnd the vertex of Lhe parabola y=2 x^{2}4 x+3. In this equation, a=2, b=− 4, and c=3. By the formula given above, the xvalue of the vertex of the parabola is x=− b2 a=− 42 (2)=1 The yvalue is found by substituting 1 for x into the equation y=2 x^{2}4 x+3 to gety=2 (1)^{2}4 (1)+3=1, so the vertex is (1,1). APPLICATION OF QUADRATIC RELATIONS Quadratic relations can be applied in situations as illustrated in the next example. PROBLEM SOLVING The fact that the vertex of a vertical parabola is the highest or lowest point on the graph makes equations of the form y=ax^{2}+bx+c important in problems where the maximum or minimum value of some quantity is to be found. When a<0, the yvalue of the vertex gives the maximum value of y and the xvalue tells where it occurs. Similarly, when a>0, the yvalue of the vertex gives the minimum yvalue. Example 8 FINDING THE VERTEX IN AN APPLICATION Ms. Whitney owns and operates Aunt Emma‘s Pie Shop. She has hired a consultant to analyze her business operations. The consultant tells her that her proﬁt P in dollars is given by P=120 xx^{2} where x is the number of units of pies that she makes. How many units of pies should be made in order lo maximize the proﬁt? What is the maximum possible proﬁt? The proﬁt relation P can be rewritten as P=− x^{2}+120 x+0, with a=− 1,b=120, and c=0. The graph of this relation will be a parabola opening downward, so that the vertex, of the form (x,P). will be the highest point on the graph. To ﬁnd the vertex, use the fact that x=− b2 a: x=− 1202 (− 1)=60 Let x=60 in the equation to ﬁnd the value of P at the vertex. P=120 xx^{2} P=120 (60)60^{2} P=3600 The vertex is (60,3600). Figure 3.23 shows the portion of the proﬁt graph located in quadrant Ι. (Why is quadrant Ι the only one of interest here?) The maximum proﬁt of $3600 occurs when 60 units of pies are made. In this case, profit increases as more and more pies are made up to 60 units and then decreases as more pies are made past this point.
Figure 3.23 In this section, we started by deﬁning a quadratic relation y=ax^{2}+bx+c and, by pointplotting, found the graph, which we called a parabola. It is possible to start with a parabola. a set of points in the plane, and ﬁnd the corresponding relation, using the formal geometric deﬁnition of a parabola. GEOMETRIC DEFINITION OF A PARABOLA Geometrically, a parabola is deﬁned as the set of all points in a plane that are equally distant from a ﬁxed point and a fixed line not containing the point. The point is called the focus and the line is the directrix. The line through the focus and perpendicular to the directrix is the axis of the parabola. The point on the axis that is equally distant from the focus and the directrix is the vertex of the parabola.
Figure 3.24 The parabola in Figure 3.24 has the point (0,p) as focus and the line y=− p as directrix. The vertex is (0,0). Let (x,y) be any point on the parabola. The distance from (x,y) to the directrix is y(− p), while the distance from (x,y) to (0,p) is √(x0)^{2}+(yp)^{2}. Since (x,y) is equally distant from the directrix and the focus, y(− p)=√(x0)^{2}+(yp)^{2} Square both sides, getting (y+p)^{2}=x^{2}+(yp)^{2} y^{2}+2 py+p^{2}=x^{2}+y^{2}2 py+p^{2} 4 py=x^{2} the equation of the parabola with focus (0,p) and directrix y=− p. Solving 4 py=x^{2} for y gives y=14 p x^{2},
Figure 3.25 so that 14 p=awhen the equation is written in the form y=ax^{2}+bx+c. This result could be extended to a parabola with vertex at (h,k), focus p units above (h,k), and directrix p units below (h,k), or to a parabola with vertex at (h,k), focus p units below (h,k), and directrix p units above (h,k). The geometric properties of parabolas lead to many practical applications. For example, if a light source is placed at the focus of a parabolic reﬂector, as in Figure 3.25, light rays reﬂect parallel to the axis, making a spotlight or ﬂashlight. The process also works in reverse. Light rays from a distant source come in parallel to the axis and are reﬂected to a point at the focus. (If such a reﬂector is aimed at the sun, a temperature of several thousand degrees may be obtained.) This use of parabolic reﬂection is seen in the satellite dishes used to pick up signals from communications satellites. HORIZONTAL PARABOLAS The directrix of a parabola could be the vertical line x=− p, where p>0. with focus on the xaxis at (p,0), producing a parabola opening to the right. This parabola is the graph of the relation y^{2}=4 px or x=[l4 p] y^{2}. The next examples show the graphs of horizontal parabolas with equations of the form x=ay^{2}+by+c. Example 9 GRAPHING A HORIZONTAL PARABOLA
Figure 2.6 Graph x=y^{2}. The equation x=y^{2} can be obtained from y=x^{2} by exchanging x and y. Choosing values of y and ﬁnding the corresponding values of x gives the parabola in Figure 3.26. The graph of x=y^{2}, shown in red, is symmetric with respect to the line y=0 and has vertex at (0,0). For comparison. the graph of y=x^{2} is shown in blue. These graphs are mirror images of each other with respect to the line y=x. From the graph, the domain of x=y^{2} is (0,∞), and the range is (− ∞,∞). NOTE The domain and the range of a horizontal parabola, such as x=y^{2} in Figure 3.26, can be determined by looking at the graph. Since the vertex (0,0) has the smallest xxvalue of any point on the graph, and the graph extends indeﬁnitely to the right. the domain is (0,∞). Because the graph extends upward and downward indeﬁnitely, the range is (− ∞,∞). EXAMPLE 10 To write this equation in the form x=a (yk)^{2}+h, complete the square on y as follows: x=2 y^{2}+6 y+5 x2y^{2}+3 y+52 Divide by 2 x252=y^{2}+3 y Subtract 52. x252+94=y^{2}+3 y+94 Add 94. x214=(y+32)^{2} Combine terms; factor. x2=(y+32)^{2}+14 Add 14. x=2 (y+32)^{2}+12 Multiply by 2. As this result shows. the vertex of the parabola is the point (12,− 32). The axis is the horizontal line y+32=0 or y=− 32.
Figure 2.7 There is no yintercept, since the vertex is on the right of the yaxis and the graph opens to the right. However, the xintercept is x=2 (0)^{2}+6 (0)+5=5. Using the vertex. the axis of symmetry, and the xintercept, and plotting a few additional points gives the graph in Figure 3.27. The domain is (12,∞) and the range is (− ∞,∞). The vertex of a horizontal parabola can also be found by using the values of a and b in x=ay^{2}+by+0.
Let’s see how our solver generates graph of this and similar problems. Click on "Solve Similar" button to see more examples. VERTEX OF A PARABOLA x=ay^{2}+by+c The yvalue of the vertex of the parabola x=ay^{2}+by+c, where a≠0, is − b2 a The xvalue is found by substitution of − b2 a for y. CAUTION Be careful when using the two vertex formulas of this section. It is essential that you recognize whether the parabola is a vertical parabola or a horizontal one, so that you can decide whether − b2 a represents the x or ycoordinate of the vertex. (It always represents the coordinate of the variable that is squared.)
