Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 13 Polynomial Pictures Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 13 Solutions Polynomial Pictures

Polynomial Pictures Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 13 Polynomial Pictures Solutions Questions and Answers

Class 9 Maths Chapter 13 Kerala Syllabus – First Degree Polynomials

Textual Questions And Answers

Question 1.
Draw the graphs of these polynomials:
i) p(x) = 2x – 1
ii) p(x) = x – 1
iii) p(x) = 1 – x
iv) p(x) = x
v) p(x) = -x
Answer:
i) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 2x – 1

x 0 1 2
P(x) -1 1 3

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 1

ii) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x – 1

x 0 1 2
P(x) -1 0 1

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 2

iii) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 1 – x

x 0 1 2
P(x) 1 0 -1

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 3

iv) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x

x 0 1 2
P(x) 0 1 2

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 4

v) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = -x

x 0 1 2
P(x) 0 -1 -2

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 5

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures

Question 2.
Find the polynomials which has these lines as their graphs:
Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 6
Answer:
i) From the figure, we get
P(0) = 1
P(\(\frac{-1}{2}\)) = 0
Since p(x) is first degree polynomial
p(x) = ax + b
p(0) = a x 0 + b = 1
b = 1
p(\(\frac{-1}{2}\)) = a x (\(\frac{-1}{2}\)) + b = 0
\(\frac{-a}{2}\) + 1 = 0
a = 2
Therefore the polynomial is 2x + 1

ii) from the figure, we get
P(0) = 0
P(\(\frac{1}{2}\)) = 1
since p(x) is first degree polynomial
p(x) = ax + b
p(0) = a × 0 + b = 0
b = 0
P(\(\frac{1}{2}\)) = a × (\(\frac{1}{2}\)) + b = 1
= \(\frac{a}{2}\) = 1
= a = 2
Therefore the polynomial is 2x

iii) from the figure, we get
P(0) = -2
P(2) = 0
since p(x) is first degree polynomial
p(x) = ax + b
p(0) = a × 0 + b = -2
b = -2
p(2) = a x (2) + b = 0
= 2a – 2 = 0
= a = 1
Therefore the polynomial is x – 2

Class 9 Maths Kerala Syllabus Chapter 13 Solutions – Second Degree Polynomials

Intext Questions And Answers

Question 1.
Draw the graphs of these polynomials:
i) p(x) = x
ii) p(x) = 2x
iii) PO) = \(\frac{1}{2}\)x
iv) p(x) = -x
v) p(x) = -2x
Do you see any common feature of these graphs?
Answer:
i) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x

x 0 1 2
P(x) 0 1 -2

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 7

ii) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 2x

x 0 1 2
P(x) 0 1 4

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 8

iii) Let’s take x = 0, x = 2, and x = 4 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = \(\frac{1}{2}\)x

x 0 2 4
P(x) 0 1 2

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 9

iv) Let’s take x = 0, x =1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = -x

x 0 1 2
P(x) 0 -1 -2

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 10

v) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = -2x

x 0 1 2
P(x) 0 -2 -4

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 11
All these lines pass through the origin

Question 2.
Now draw the graphs of these polynomials:
i) p(x) = x +1
ii) p(x) = x + 2
iii) P(x) = x + \(\frac{1}{2}\)
iv) p(x) = x – 1
v) p(x) = x – 2
Do you see any common feature of these graphs?
Answer:
i) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x + 1

x 0 1 2
P(x) 1 2 3

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 12

ii) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x + 2

x 0 1 2
P(x) 2 3 4

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 13

iii) Let’s take x = 0, x = \(\frac{1}{2}\), and x = 1 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x + \(\frac{1}{2}\)

x 0 \(\frac{1}{2}\) 1
P(x) \(\frac{1}{2}\) 1 1\(\frac{1}{2}\)

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 14

iv) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x – 1

x 0 1 2
P(x) -1 0 1

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 15

v) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = x – 2

x 0 1 2
P(x) -2 -1 0

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 16
All these lines have same slope

Polynomial Pictures Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
Draw the graphs of these polynomials:
i) P(x) = 3 – x
ii) p(x) = 3x + 1
iii) p(x) = 2x – 3
Answer:
i) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 3 – x

x 0 1 2
P(x) 3 2 1

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 17

ii) Lets take x = -1, x = 0, and x = 1 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 3x + 1

x -1 0 1
P(x) -2 1 4

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 18

iii) Let’s take x = 0, x = 1, and x = 2 and draw the table to find out the corresponding p(x) values for the polynomial p(x) = 2x – 3

x 0 1 2
P(x) -3 -1 1

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 19

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures

Question 2.
Find the polynomials which has these lines as their graphs:
Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 20
Answer:
i) from figure we get,
P(0) = 8
p(-4) = 0
since p(x) is first degree polynomial
now, p(x) = ax + b
p(0) = a × 0 + b = 8
b = 8

p(-4) = a × (-4) + b = 0
-4a + 8 = 0
a = 2
Therefore, the polynomial is p(x) = 2x + 8

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 21
Answer:
from figure we get,
p(0) = -4
p(4) = 0
since p(x) is first degree polynomial now,
p(x) = ax + b
p(0) = a × 0 + b = -4
b = -4
p(4) = a × (4) + b = 0
= 4a – 4 = 0
= a = 1
Therefore, the polynomial is p(x) = x – 4

Question 3.
The graphs of some second degree polynomials are given below:
Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 22
Answer:
from figure we get, p(0) = 1
p(-2) = 5
P(2) = 5
since p(x) is first degree polynomial
p(x) = ax² + bx + c
now, p(0) = a × 0² + b × 0 + c = 1
= c = 1

p(-2) = a × (-2)² + bx – 2 + c = 5
= 4a – 2b + 1 = 5
= 4a- 2b = 4
= 2a-b = 2 …(1)

p(2) = a × 2² + bx² + c = 0
= 4a + 2b + 1 = 5
= 4a + 2b = 4
= 2a + b = 2 … (2)

Adding equation (1) and equation (2)
4a = 4
a = 1

Put a = 1 in equation (1) we get,
2 × 1 – b = 2
b = 0
Therefore, the polynomial is p(x) = x² + 1

Kerala Syllabus Class 9 Maths Chapter 13 Solutions Polynomial Pictures 23
Answer:
from the figure we get,
P(0) = -2
p(-2) = 2
p(2) = 2
since p(x) is second degree polynomial p(x) = ax² + bx + c
now, p(0) = a × 0² + b x 0 + c = -2
= c = -2

p(-2) = a × (-2)² + bx – 2 + c = 2
= 4a – 2b – 2 = 2
= 4a – 2b = 4
= 2a – b = 2 …(1)

p(2) = a × 2² + bx² + c = 0
= 4a + 2b – 2
= 2

Adding equation (1) and equation (2)
4a = 4
a = 5 = 1

Put a = 1 in equation (1) we get,
2 × 1 – b = 2
b = 0
Therefore, the polynomial is p(x) = x² – 2

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