The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11). Class 9

The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11). Class 9

Question 1.

The graph of a linear polynomial p(x) passes through the points (1, 5) and (3, 11). Class 9

i. Find the polynomial p(x).

ii. Find the coordinates where the graph of p(x) cuts the axes.

iii. Draw the graph of p(x) and verify your answers.

Solution:

Let the linear polynomial be

p(x) = ax + b ...(i)

Since the graph of a linear polynomial passes through (1, 5) and (3, 11)

∴ Both points satisfy p(x)

By using point (1, 5), we get

∴ a + b = 5 ...(ii)

3a + b = ll ...(iii)

From (ii), we get

a = 5 - b .....(iv)

Substituting a = 5 - b in (iii), we get

3(5 - b) + b = 11

∴ 15 - 3b + b = 11

∴ 15 - 2b = 11

∴ -2b = 11 - 15

∴ -2b = -4

∴ b = 2

Now, Substituting b = 2 in (iv), we get

a = 5 - 2 = 3

Substituting a = 3 and b = 2 in (i), we get

p(x) = 3x + 2

ii. When p(x) cuts the y-axis,

x = 0

∴ p (0) = 3(0) + 2 = 2

∴ The graph cuts the y-axis at (0, 2).

When p(x) cuts the x-axis,

p(x) = 0

∴ 0 = 3x + 2

∴ 3x = - 2

∴ x = $-\frac{2}{3}$

∴ The graph cuts the x-axis at ($-\frac{2}{3}$, 0)

For a linear equation y = ax + b

When x = 0, the line cuts y - axis

When y = 0, the line cuts x - axis

iii.

From the graph, we observe that p(x) cuts the x-axis and y-axis as ($-\frac{2}{3}$, 0) and (0, 2) respectively.