2.3 Redes inal´ambricas de sensores
3.1.2 Software empotrado
A general quadratic equation is of the form y=ax2+bx+c, wherea,b andcare constants and ais not equal to zero.
A graph of a quadratic equation always produces a shape called aparabola.
The gradients of the curves between 0 and A and between B and C in Figure 19.3 are positive, whilst the gradient betweenAandBis negative. Points such as AandBare calledturning points. At Athe gradi-ent is zero and, asxincreases, the gradient of the curve changes from positive just beforeAto negative just after.
Such a point is called amaximum value. AtBthe gra-dient is also zero and, asxincreases, the gradient of the curve changes from negative just before B to positive just after. Such a point is called aminimum value.
y
x
A C
B 0
Figure 19.3
Following are three examples of solutions using quadratic graphs.
(a) y=ax2
Graphs of y=x2,y=3x2 and y=1 2x2 are shown in Figure 19.4. All have minimum values at the origin (0, 0).
Graphical solution of equations 157
(a) y
y5x2
x 2 1 21 0 1
(b) y53x2 y
x 2 1 21 0 1
y
x 2 1 21 0 1
(c)
y5 x21 2
Figure 19.4
Graphs ofy= −x2,y= −3x2andy= −1 2x2are shown in Figure 19.5. All have maximum values at the origin (0, 0).
(a) y
y52x2 21 x
21 22 0
1
(b) y523x2 21
22 y 21 x
0 1
21 22 y 21 0 x
1
(c)
y52 x21 2
Figure 19.5
Wheny=ax2,
(i) curves are symmetrical about they-axis, (ii) the magnitude ofaaffects the gradient of the
curve, and
(iii) the sign of a determines whether it has a maximum or minimum value.
(b) y=ax2+c
Graphs of y=x2+3,y=x2−2,y= −x2+2 andy= −2x2−1 are shown in Figure 19.6.
Wheny=ax2+c,
(i) curves are symmetrical about they-axis, (ii) the magnitude ofaaffects the gradient of the
curve, and
(iii) the constantcis they-axis intercept.
(c) y=ax2+bx+c
Wheneverbhas a value other than zero the curve is displaced to the right or left of they-axis.
Whenb/ais positive, the curve is displacedb/2a to the left of they-axis, as shown in Figure 19.7(a).
When b/a is negative, the curve is displaced b/2a to the right of the y-axis, as shown in Figure 19.7(b).
Quadratic equations of the form ax2+bx+c=0 may be solved graphically by
(a) plotting the graph y=ax2+bx+c, and
(b) noting the points of intersection on thex-axis (i.e.
wherey=0).
y
y5x213
y5 2x212
y5 22x221 y5x222
21 0 x (a)
3
1
y
x 22
21 1
(b) 2 0
y
21 x
(c) 0
2
1
y 21 x
21 24
(d) 0
1
Figure 19.6
(a)
2524 23 22 21 1 y
x 12 10
6 4 2 0 y5x216x111
(b) y5x225x14
21 y
0 x 2 4 6
22
1 2 3 4
8
Figure 19.7
The x values of the points of intersection give the required solutions since at these points bothy=0 and ax2+bx+c=0.
The number of solutions, or roots, of a quadratic equa-tion depends on how many times the curve cuts the x-axis. There can be no real roots, as in Figure 19.7(a), one root, as in Figures 19.4 and 19.5, or two roots, as in Figure 19.7(b).
Here are some worked problems to demonstrate the graphical solution of quadratic equations.
Problem 3. Solve the quadratic equation 4x2+4x−15=0 graphically, given that the solutions lie in the rangex= −3 tox=2.
Determine also the co-ordinates and nature of the turning point of the curve
158 Basic Engineering Mathematics
Lety=4x2+4x−15. A table of values is drawn up as shown below.
x −3 −2 −1 0 1 2
y=4x2+4x−15 9 −7 −15 −15 −7 9
y54x214x215 12
y
8 4
24 28 212
216 20.5
1.5
0 1 2 x
A B
22.5
23 22 21
Figure 19.8
A graph ofy=4x2+4x−15 is shown in Figure 19.8.
The only points wherey=4x2+4x−15 andy=0 are the points marked A andB. This occurs at x= −2.5 andx=1.5and these are the solutions of the quadratic equation 4x2+4x−15=0.
By substitutingx= −2.5 andx=1.5 into the original equation the solutions may be checked.
The curve has a turning point at(−0.5,−16)and the nature of the point is aminimum.
An alternative graphical method of solving 4x2+4x−15=0 is to rearrange the equation as 4x2= −4x+15 and then plot two separate graphs
− in this case, y=4x2 and y= −4x+15. Their points of intersection give the roots of the equation 4x2= −4x+15, i.e. 4x2+4x−15=0. This is shown in Figure 19.9, where the roots are x= −2.5 and x=1.5, as before.
Problem 4. Solve graphically the quadratic equation−5x2+9x+7.2=0 given that the solutions lie betweenx= −1 andx=3. Determine also the co-ordinates of the turning point and state its nature
y5 24x115 30
25 20 15 10 5 y
0 1 2 3 x
21 22
22.5 1.5
23
y54x2
Figure 19.9
Lety= −5x2+9x+7.2. A table of values is drawn up as shown below.
x −1 −0.5 0 1
y= −5x2+9x+7.2 −6.8 1.45 7.2 11.2
x 2 2.5 3
y= −5x2+9x+7.2 5.2 −1.55 −10.8 A graph of y= −5x2+9x+7.2 is shown plotted in Figure 19.10. The graph crosses thex-axis (i.e. where y=0) atx= −0.6andx=2.4and these are the solu-tions of the quadratic equation−5x2+9x+7.2=0.
The turning point is amaximum, having co-ordinates (0.9,11.25).
Problem 5. Plot a graph ofy=2x2and hence solve the equations
(a) 2x2−8=0 (b) 2x2−x−3=0 A graph ofy=2x2is shown in Figure 19.11.
(a) Rearranging 2x2−8=0 gives 2x2=8 and the solution of this equation is obtained from the points of intersection of y=2x2 and y=8; i.e., at co-ordinates(−2,8)and(2,8), shown as Aand B, respectively, in Figure 19.11. Hence, the solutions of 2x2−8=0 arex= −2andx= +2.
(b) Rearranging 2x2−x−3=0 gives 2x2=x+3 and the solution of this equation is obtained from the points of intersection of y=2x2 and y=x+3; i.e., atCandDin Figure 19.11. Hence, the solutions of 2x2−x−3=0 arex= −1and x=1.5
Graphical solution of equations 159
⫺10
⫺8
⫺6
⫺4
⫺0.6⫺2
y⫽ ⫺5x2⫹9x⫹7.2
0.9 2.4
⫺1 0 1 2 3
2 4 6 8 10 11.2512 y
x
Figure 19.10
y⫽2x2
y⫽8
y⫽x⫹3
⫺2 ⫺1 0 2 4 6 8 10 A
C
D B y
1 1.5 2 x
Figure 19.11
Problem 6. Plot the graph ofy= −2x2+3x+6 for values ofxfromx= −2 tox=4. Use the graph to find the roots of the following equations.
(a) −2x2+3x+6=0 (b) −2x2+3x+2=0 (c) −2x2+3x+9=0 (d) −2x2+x+5=0 A table of values fory= −2x2+3x+6 is drawn up as shown below.
x −2 −1 0 1 2 3 4
y −8 1 6 7 4 −3 −14
A graph of y= −2x2+3x+6 is shown in Figure 19.12.
28 26 24 22 21.13 21.35
22 21.5 G
A B
D C
H
F y5 23 y54
y52x11 y5 22x213x16
x
E
1 2
1.85 2.63 3 0
2 4 6 8 y
20.5 21
Figure 19.12
(a) The parabolay= −2x2+3x+6 and the straight liney=0 intersect atAandB, wherex= −1.13 and x=2.63, and these are the roots of the equation−2x2+3x+6=0.
(b) Comparing y= −2x2+3x+6 (1) with 0= −2x2+3x+2 (2) shows that, if 4 is added to both sides of equation (2), the RHS of both equations will be the same. Hence, 4= −2x2+3x+6. The solution of this equation is found from the points of intersection of the line y=4 and the parabola y= −2x2+3x+6; i.e., points C and D in Figure 19.12. Hence, the roots of
−2x2+3x+2=0 arex= −0.5andx=2.
(c) −2x2+3x+9=0 may be rearranged as
−2x2+3x+6= −3 and the solution of this equation is obtained from the points of intersection of the line y= −3 and the parabola y= −2x2+3x+6; i.e., at points E and F in Figure 19.12. Hence, the roots of
−2x2+3x+9=0 arex= −1.5andx=3.
160 Basic Engineering Mathematics
(d) Comparing y= −2x2+3x+6 (3) with 0= −2x2+x+5 (4) shows that, if 2x+1 is added to both sides of equation (4), the RHS of both equations will be the same. Hence, equation (4) may be written as 2x+1= −2x2+3x+6. The solu-tion of this equasolu-tion is found from the points of intersection of the line y=2x+1 and the parabola y= −2x2+3x+6; i.e., points G and H in Figure 19.12. Hence, the roots of
−2x2+x+5=0 arex= −1.35andx=1.85
Now try the following Practice Exercise Practice Exercise 73 Solving quadratic equations graphically (answers on page 348)
1. Sketch the following graphs and state the nature and co-ordinates of their respective turning points.
(a) y=4x2 (b) y=2x2−1 (c) y= −x2+3 (d) y= −1
2x2−1 Solve graphically the quadratic equations in prob-lems 2 to 5 by plotting the curves between the given limits. Give answers correct to 1 decimal place.
2. 4x2−x−1=0; x= −1 tox=1 3. x2−3x=27; x= −5 tox=8 4. 2x2−6x−9=0; x= −2 tox=5 5. 2x(5x−2)=39.6; x= −2 tox=3 6. Solve the quadratic equation
2x2+7x+6=0 graphically, given that the solutions lie in the range x= −3 to x=1. Determine also the nature and co-ordinates of its turning point.
7. Solve graphically the quadratic equation 10x2−9x−11.2=0, given that the roots lie betweenx= −1 andx=2.
8. Plot a graph of y=3x2and hence solve the following equations.
(a) 3x2−8=0 (b) 3x2−2x−1=0 9. Plot the graphs y=2x2 andy=3−4x on the same axes and find the co-ordinates of the points of intersection. Hence, determine the roots of the equation 2x2+4x−3=0.
10. Plot a graph of y=10x2−13x−30 for values of x between x= −2 and x=3.
Solve the equation 10x2−13x−30=0 and from the graph determine
(a) the value ofywhenxis 1.3, (b) the value ofxwhenyis 10, (c) the roots of the equation
10x2−15x−18=0.