The function can be classified based on the degree of the polynomial.
Degree of the polynomial is the highest exponent of the polynomial.
Degree 0 1 2 |
Name of the polynomial Constant Linear Quadratic |
If the given polynomial is in factored form, use distributive property then do the possible simplification.
Then we can find out the degree and classify it.
Determine whether each function is linear or quadratic. Identify the quadratic, linear, and constant terms.
Problem 1 :
y = (x – 2)(x + 4)
Solution :
It is in factored form.
y = x2 + 4x - 2x - 8
y = x2 + 2x - 8
The degree of the polynomial is 2. So, it is quadratic.
Problem 2 :
y = 3x(x + 5)
Solution :
y = 3x(x + 5)
Distributing 3x, we get
y = 3x2 + 15x
Degree of the polynomial is 2. So, it is quadratic polynomial.
Problem 3 :
y = 5x(x – 5) –5x2
Solution :
y = 5x(x – 5) –5x2
y = 5x2 – 5x –5x2
By combining the like terms, we get
y = – 5x
Degree of this polynomial is 1. So, it is a linear.
Problem 4 :
f(x) = 7(x – 2) + 5(3x)
Solution :
f(x) = 7(x – 2) + 5(3x)
f(x) = 7x - 14 + 15x
By combining the like terms
f(x) = 22x - 14
The degree of the polynomial is 1. So it is linear.
Problem 5 :
f(x) = 3x2 – (4x – 8)
Solution :
f(x) = 3x2 – (4x – 8)
f(x) = 3x2 – 4x + 8
Degree of the f(x) is 2. So, it is quadratic polynomial.
Problem 6 :
y = 3x(x – 1) – (3x + 7)
Solution :
y = 3x(x – 1) – (3x + 7)
y = 3x2 - 3x - 3x - 7
By combining the like terms, we get
y = 3x2 - 6x - 7
The degree of the polynomial is 2. So, it is quadratic.
Problem 7 :
y = 3x2 – 12
Solution :
The degree of the polynomial is 2. So, it is quadratic.
Problem 8 :
f(x) = (2x – 3)(x + 2)
Solution :
The degree of the polynomial is 2. So, it is quadratic.
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