Graphs

Let X'OX and YOY' be the coordinate axes.

(i) On the x−axis, take 4 units to the right of the y axis; and then on the y−axis, take 3 units above the x−axis. Thus, we obtain the point A(4,3)

(ii) On the x−axis, take 2 units to the right of the y−axis; and then on the y−axis, take 6 units above the x−axis. Thus, we obtain the point B(2,6)

(iii) On the x−axis, take 3 units to the left of the y−axis; and then on the y−axis, take 5 units above the x−axis. Thus, we obtain the point C(−3,5)

(iv) On the x−axis, take 5 units to the left of the y−axis; and then on the y−axis, take 2 units above the x−axis. Thus, we obtain the point D(−5,2)

(v) On the x−axis, take 2 units to the left of the y−axis; and then on the y−axis, take 3 units below the x−axis. Thus, we obtain the point E(−2,−3)

(vi) On the x−axis, take 5 units to the left of the y−axis; and then on the y−axis, take 3 units below the x−axis. Thus, we obtain the point F(−5,−3)

(vii) On the x−axis, take 5 units to the right of the y−axis; and then on the y−axis, take 4 units below the x−axis. Thus, we obtain the point G(5,−4)

(viii) On the x−axis, take 3 units to the right of the y−axis; and then on the y−axis, take 3 units below the x−axis. Thus, we obtain the point H(3,−3)

The given function is y=3x. For some different values of x, the corresponding values of y are given below:

Now, let us plot the points O(0,0), A(1,3) and B(2,6).

∴ Now, we obtain our required graph.

(i) Our point C to be plotted lies on function y = 3x.

Here, first plotting y = 3x.

Here, x = 3.

∴ Now for abscissa equal to 3, we plot the point on y = 3x, ie y = 3 × 3 = 9

Hence, the value of y is 9

(ii) Our point to be plotted lies on function y = 3x.

Here, first plotting y = 3x.

Here, x = 5.

∴ Now for abscissa equal to 5, we plot the point on y = 3x, ie y = 3 × 5 = 15

Hence, the value of y is 15

(iii) Our point to be plotted lies on function y = 3x.

∴Here, first plotting y = 3x.

Here, x = 6.

∴ Now for abscissa equal to 6, we plot the point on y = 3x, ie y = 3 × 6 = 18

Hence, the value of y is 18

The given function is P = 4x. For some different values of x, the corresponding values of P are given below:

Now let us plot the points, O(0,0), A(1,4) and B(2,8)

∴ Now, we obtain our required graph.

(i) Our point C to be plotted lies on function P = 4x.

∴Here, first plotting P = 4x.

Here, x = 3.

∴ Now for abscissa equal to 3, we plot the point on P = 4x, ie P = 4 × 3 = 12

Hence, the value of P is 12

(ii) Our point D to be plotted lies on function P = 4x.

∴Here, first plotting P = 4x.

Here, x = 4.

∴ Now for abscissa equal to 4, we plot the point on P = 4x, ie P = 4 × 4 = 16

Hence, the value of P is 16

(iii) Our point E to be plotted lies on function P = 4x.

∴Here, first plotting P = 4x.

Here, x = 6.

∴ Now for abscissa equal to 6, we plot the point on P = 4x, ie P = 4 × 6 = 24

Hence, the value of P is 24

The given function is A=x² .

For some different values of x, the corresponding values of A are given below:

Now let us plot the points, O(0,0), S(1,1) and P(2,4).

∴ Now we obtain the required graph.

(i) Our point B to be plotted lies on function A = x².

∴Here, first plotting A = x².

Here, x = 2.

∴ Now for abscissa equal to 2, we plot the point on A = x²., ie A = 2² = 4

Hence, the value of A is 4

(ii) Our point C to be plotted lies on function A = x².

∴Here, first plotting A = x².

Here, x = 3.

∴ Now for abscissa equal to 3, we plot the point on A = x²., ie A = 3² = 9

Hence, the value of A is 9

(iii) Our point to be plotted lies on function A = x².

∴Here, first plotting A = x².

Here, x = 4.

∴ Now for abscissa equal to 4, we plot the point on A = x²., ie A = 4² = 16

Hence, the value of A is 16

Here, given point is P(3,6).

Both the coordinates are positive.

Hence, point P lies in first quadrant.

Here, given point is (-7,-1).

Both the coordinates are negative.

Hence, given point lies in third quadrant.

Here, given point is .

Here, abscissa of a point is positive and ordinate is negative.

Hence, given point lies in fourth quadrant.

Here, given point is

Here, abscissa of a point is negative and ordinate is positive.

Hence, given point lies in second quadrant.

We know that the abscissa of a point is its distance from the y−axis.

Here, the line y = a is parallel x-axis.

We know that the graph x = a is a line parallel to the y-axis.

Hence, for x = 0, line represents y axis.