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 |
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 |
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 |
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 |
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 |
Question 2.
Find the polynomials which has these lines as their graphs:
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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\(\frac{1}{2}\) |
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 |
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 |
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 |
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 |
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 |
Question 2.
Find the polynomials which has these lines as their graphs:
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
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:
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
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