4.6 Emparejamiento óptimo ”ate“ para curvas BN
5.1.2 Las mejores implementaciones con curvas ordinarias
y=mx+c
wheremis the gradient andcis they-axis intercept.
Thus, as we have found in Figure 17.8(a), y=2x+1 represents a straight line of gradient 2 and y-axis intercept 1. So, given the equation y=2x+1, we are able to state, on sight, that the gradient =2 and they-axis intercept=1, without the need for any analysis.
Similarly, in Figure 17.8(b),y= −3x+2 represents a straight line of gradient−3 andy-axis intercept 2.
In Figure 17.8(c),y=3 may be rewritten asy=0x+3 and therefore represents a straight line of gradient 0 and y-axis intercept 3.
Here are some worked problems to help understanding of gradients, intercepts and equations of graphs.
Problem 2. Plot the following graphs on the same axes in the rangex= −4 tox= +4 and determine the gradient of each.
(a) y=x (b)y=x+2
(c) y=x+5 (d)y=x−3
A table of co-ordinates is produced for each graph.
(a) y=x
x −4 −3 −2 −1 0 1 2 3 4 y −4 −3 −2 −1 0 1 2 3 4 (b) y=x+2
x −4 −3 −2 −1 0 1 2 3 4 y −2 −1 0 1 2 3 4 5 6 (c) y=x+5
x −4 −3 −2 −1 0 1 2 3 4
y 1 2 3 4 5 6 7 8 9
(d) y=x−3
x −4 −3 −2 −1 0 1 2 3 4
y −7 −6 −5 −4 −3 −2 −1 0 1 The co-ordinates are plotted and joined for each graph.
The results are shown in Figure 17.9. Each of the straight lines produced is parallel to the others; i.e., the slope or gradient is the same for each.
136 Basic Engineering Mathematics
⫺4⫺3⫺2⫺1 1 2 3 4 x A
B C
F E D y⫽x⫹5
y⫽x⫹2
y⫽x⫺3 y⫽x 9
y 8 7 6 5 4 3 2 1
⫺1
⫺2
⫺3
⫺4
⫺5
⫺6
⫺7 Figure 17.9
To find the gradient of any straight line, say,y=x−3, a horizontal and vertical component needs to be con-structed. In Figure 17.9,ABis constructed vertically at x=4 andBCis constructed horizontally aty= −3.
The gradient of AC=AB
BC =1−(−3) 4−0 =4
4=1 i.e. the gradient of the straight liney=x−3 is 1, which could have been deduced ‘on sight’ since y=1x−3 represents a straight line graph with gradient 1 and y-axis intercept of−3.
The actual positioning of AB andBC is unimportant because the gradient is also given by
DE
EF =−1−(−2) 2−1 =1
1 =1
The slope or gradient of each of the straight lines in Figure 17.9 is thus 1 since they are parallel to each other.
Problem 3. Plot the following graphs on the same axes between the valuesx= −3 tox= +3 and determine the gradient andy-axis intercept of each.
(a)y=3x (b)y=3x+7
(c)y= −4x+4 (d)y= −4x−5
A table of co-ordinates is drawn up for each equation.
(a) y=3x
x −3 −2 −1 0 1 2 3 y −9 −6 −3 0 3 6 9 (b) y=3x+7
x −3 −2 −1 0 1 2 3
y −2 1 4 7 10 13 16
(c) y= −4x+4
x −3 −2 −1 0 1 2 3
y 16 12 8 4 0 −4 −8
(d) y= −4x−5
x −3 −2 −1 0 1 2 3
y 7 3 −1 −5 −9 −13 −17 Each of the graphs is plotted as shown in Figure 17.10 and each is a straight line.y=3x andy=3x+7 are parallel to each other and thus have the same gradient.
The gradient ofACis given by CB
BA =16−7 3−0 =9
3 =3
y53x17 y52
4x2 5 y52
4x1 4
y53x 16
12 8 4
23 22 0
28 212 216
21 1 2 3 x
24
B C
A
F
E D
y
Figure 17.10
Hence,the gradients of bothy=3xandy=3x+7 are 3, which could have been deduced ‘on sight’.
y= −4x+4 andy= −4x−5 are parallel to each other and thus have the same gradient. The gradient ofDFis
Straight line graphs 137
given by FE
ED=−5−(−17) 0−3 = 12
−3= −4
Hence, the gradient of both y= −4x+4 and y= −4x−5 is −4, which, again, could have been deduced ‘on sight’.
The y-axis intercept means the value of y where the straight line cuts they-axis. From Figure 17.10,
y=3xcuts they-axis aty=0 y=3x+7 cuts they-axis aty= +7 y= −4x+4 cuts they-axis aty= +4 y= −4x−5 cuts they-axis aty= −5
Some general conclusions can be drawn from the graphs shown in Figures 17.9 and 17.10. When an equation is of the formy=mx+c, wherem andcare constants, then
(a) agraph ofyagainstxproduces a straight line, (b) mrepresents the slope or gradient of the line, and (c) crepresents they-axis intercept.
Thus, given an equation such asy=3x+7, it may be deduced ‘on sight’ that its gradient is+3 and itsy-axis intercept is+7, as shown in Figure 17.10. Similarly, if y= −4x−5, the gradient is−4 and they-axis intercept is−5, as shown in Figure 17.10.
When plotting a graph of the formy=mx+c, only two co-ordinates need be determined. When the co-ordinates are plotted a straight line is drawn between the two points. Normally, three co-ordinates are determined, the third one acting as a check.
Problem 4. Plot the graph 3x+y+1=0 and 2y−5=xon the same axes and find their point of intersection
Rearranging 3x+y+1=0 gives y= −3x−1 Rearranging 2y−5=x gives 2y=x+5 and
y=1 2x+21
2 Since both equations are of the formy=mx+c, both are straight lines. Knowing an equation is a straight line means that only two co-ordinates need to be plot-ted and a straight line drawn through them. A third co-ordinate is usually determined to act as a check.
A table of values is produced for each equation as shown below.
x 1 0 −1
−3x−1 −4 −1 2
x 2 0 −3
1
2x+212 312 212 1 The graphs are plotted as shown in Figure 17.11.
Thetwo straight lines are seen to intersect at(−1,2).
4
24 23 22 21 0 3 2 1
21 22 23 24 y5 23x21
y
x
1 2 3 4
y5 12x1
5 2
Figure 17.11
Problem 5. If graphs ofyagainstxwere to be plotted for each of the following, state (i) the gradient and (ii) they-axis intercept.
(a) y=9x+2 (b)y= −4x+7 (c) y=3x (d)y= −5x−3
(e) y=6 (f )y=x
If y=mx+c then m=gradient and c=y-axis intercept.
(a) Ify=9x+2, then (i) gradient=9 (ii)y-axis intercept=2 (b) Ify= −4x+7, then (i)gradient= −4
(ii)y-axis intercept=7 (c) Ify=3xi.e.y=3x+0,
then (i) gradient=3
(ii)y-axis intercept=0 i.e. the straight line passes through the origin.
138 Basic Engineering Mathematics
(d) Ify= −5x−3, then (i)gradient= −5 (ii)y-axis intercept= −3 (e) If y=6 i.e. y=0x+6,
then (i)gradient=0
(ii)y-axis intercept=6 i.e.y=6 is a straight horizontal line.
(f ) If y=x i.e. y=1x+0,
then (i)gradient=1
(ii)y-axis intercept=0 Sincey=x, asxincreases,yincreases by the same amount; i.e.,yis directly proportional tox.
Problem 6. Without drawing graphs, determine the gradient andy-axis intercept for each of the following equations.
(a) y+4=3x (b) 2y+8x=6 (c) 3x=4y+7 If y=mx+c then m=gradient and c=y-axis intercept.
(a) Transposingy+4=3xgives y=3x−4 Hence,gradient =3andy-axis intercept= −4.
(b) Transposing 2y+8x=6 gives 2y= −8x+6 Dividing both sides by 2 gives y= −4x+3 Hence,gradient= −4andy-axis intercept=3.
(c) Transposing 3x=4y+7 gives 3x−7=4y
or 4y=3x−7
Dividing both sides by 4 gives y=3 4x−7
4
or y=0.75x−1.75
Hence, gradient = 0.75 and y-axis intercept
= −1.75
Problem 7. Without plotting graphs, determine the gradient andy-axis intercept values of the following equations.
(a)y=7x−3 (b) 3y= −6x+2 (c)y−2=4x+9 (d) y
3 =x 3−1
5 (e) 2x+9y+1=0
(a) y=7x−3 is of the formy=mx+c
Hence, gradient, m=7 and y-axis intercept, c= −3.
(b) Rearranging 3y= −6x+2 gives y= −6x 3 +2
3, i.e. y= −2x+2 which is of the formy=mx+c 3
Hence,gradient m= −2 andy-axis intercept, c=2
3
(c) Rearrangingy−2=4x+9 givesy=4x+11.
Hence,gradient=4andy-axis intercept=11.
(d) Rearranging y 3=x
2−1
5 gives y=3 x
2 −1 5
=3 2x−3
5 Hence,gradient=3
2 and y-axis intercept= −3
5
(e) Rearranging 2x+9y+1=0 gives 9y= −2x−1, i.e. y= −2
9x−1 9 Hence,gradient= −2
9 and y-axis intercept= −1
9
Problem 8. Determine for the straight line shown in Figure 17.12 (a) the gradient and (b) the equation of the graph
23 20
24 23 22 0 15 10 8 5
25 210 215 220 y
x
1 2 3 4
21
Figure 17.12
(a) A right-angled triangleABCis constructed on the graph as shown in Figure 17.13.
Gradient=AC
CB =23−8 4−1 =15
3 =5
Straight line graphs 139
23 20
24 23 22 0 15 10 8 5
25 210 215 220 y
x C B
A
1 2 3 4
21
Figure 17.13
(b) They-axis intercept atx=0 is seen to be aty=3.
y=mx+c is a straight line graph where m=gradient andc=y-axis intercept.
From above,m=5 andc=3.
Hence, the equation of the graph isy=5x+3.
Problem 9. Determine the equation of the straight line shown in Figure 17.14.
4
24 23 22 0 3 2 1
21 22 23 24 y
x
F E D
1 2 3 4
21
Figure 17.14
The triangleDEFis shown constructed in Figure 17.14.
Gradient ofDE=DF
FE =3−(−3)
−1−2 = 6
−3 = −2 and they-axis intercept=1.
Hence,the equation of the straight line isy=mx+c i.e.y= −2x+1.
Problem 10. The velocity of a body was measured at various times and the results obtained were as follows:
Velocityv(m/s) 8 10.5 13 15.5 18 20.5 23
Timet(s) 1 2 3 4 5 6 7
Plot a graph of velocity (vertically) against time (horizontally) and determine the equation of the graph
Suitable scales are chosen and the co-ordinates (1, 8), (2, 10.5), (3, 13), and so on, are plotted as shown in Figure 17.15.
0 4 5.56
8 Q R
P
10 12 14 16
Velocity (y), in metres per second
18 20 22
1 2 3
Time (t), in seconds
4 5 6 7
Figure 17.15
The right-angled triangle PRQ is constructed on the graph as shown in Figure 17.15.
Gradient ofPQ=PR
RQ=18−8 5−1 =10
4 =2.5 Thevertical axis interceptis atv=5.5 m/s.
The equation of a straight line graph isy=mx+c. In this case, t corresponds tox andv corresponds to y.
Hence, the equation of the graph shown in Figure 17.15
140 Basic Engineering Mathematics
isv=mt+c. But, from above, gradient,m=2.5 and v-axis intercept,c=5.5.
Hence,the equation of the graph isv=2.5t+5.5
Problem 11. Determine the gradient of the straight line graph passing through the co-ordinates (a)(−2,5)and(3,4), and (b)(−2,−3)and(−1,3) From Figure 17.16, a straight line graph passing through co-ordinates(x1,y1)and(x2,y2)has a gradient given by
m= y2−y1
x2−x1
(x1, y1)
(x2, y2)
(x2⫺x1)
(y2⫺ y1)
0 y y2
y1
x1 x2 x
Figure 17.16
(a) A straight line passes through(−2,5)and(3,4), hencex1= −2,y1=5,x2=3 andy2=4, hence, gradient,m= y2−y1
x2−x1 = 4−5
3−(−2)=−1 5 (b) A straight line passes through (−2,−3) and
(−1,3), hence x1= −2,y1= −3, x2= −1 and y2=3, hence,gradient,
m= y2−y1
x2−x1 = 3−(−3)
−1−(−2)= 3+3
−1+2 =6 1=6
Now try the following Practice Exercise
Practice Exercise 68 Gradients, intercepts and equations of graphs (answers on page 347)
1. The equation of a line is 4y=2x+5. A table of corresponding values is produced and is shown below. Complete the table and plot a graph ofyagainstx. Find the gradient of the graph.
x −4 −3 −2 −1 0 1 2 3 4
y −0.25 1.25 3.25 2. Determine the gradient and intercept on the y-axis for each of the following equations.
(a) y=4x−2 (b) y= −x (c) y= −3x−4 (d) y=4
3. Find the gradient and intercept on they-axis for each of the following equations.
(a) 2y−1=4x (b) 6x−2y=5 (c) 3(2y−1)=x
4
Determine the gradient andy-axis intercept for each of the equations in problems 4 and 5 and sketch the graphs.
4. (a)y=6x−3 (b)y= −2x+4 (c)y=3x (d)y=7
5. (a) 2y+1=4x (b) 2x+3y+5=0 (c) 3(2y−4)=x
3 (d) 5x−y 2−7
3=0 6. Determine the gradient of the straight line
graphs passing through the co-ordinates:
(a) (2, 7) and(−3,4) (b) (−4,−1)and(−5,3) (c)
1 4,−3
4
and
−1 2,5
8
7. State which of the following equations will produce graphs which are parallel to one another.
(a) y−4=2x (b) 4x= −(y+1) (c) x=1
2(y+5) (d) 1+1 2y=3
2x (e) 2x=1
2(7−y)
8. Draw on the same axes the graphs of y=3x−5 and 3y+2x=7. Find the co-ordinates of the point of intersection. Check the result obtained by solving the two simul-taneous equations algebraically.
9. Plot the graphsy=2x+3 and 2y=15−2x on the same axes and determine their point of intersection.
10. Draw on the same axes the graphs of y=3x−1 and y+2x=4. Find the co-ordinates of the point of intersection.