To check if the given ordered pair is a solution, we have to follow the given steps.
Step 1 :
Consider the given solution ordered pair as (x, y)
Step 2 :
Apply the values of x and y, in the given quadratic function.
Step 3 :
State whether the following quadratic functions are satisfied by the given ordered pairs
Problem 1 :
y = 3x2 + 2x (2, 16)
Solution :
y = 3x2 + 2x
Given ordered pair is (2, 16)
16 = 3(2)2 + 2(2)
16 = 3(4) + 4
16 = 12 + 4
16 = 16
Since the given ordered pair is satisfying the given quadratic function, it is a solution.
Problem 2:
f(x) = -x2 - 2x + 1 (-3, 1)
Solution :
f(x) = -x2 - 2x + 1
Given ordered pair is (-3, 1)
f(-3) = -(-3)2 - 2(-3) + 1
1 = -9 + 6 + 1
1 = -9 + 7
1 ≠ -2
Since the given ordered pair is not satisfying the given quadratic function, it is not a solution.
Problem 3 :
f(x) = 5x2 - 10 (0, 5)
Solution :
f(x) = 5x2 - 10
Given ordered pair is (0, 5)
f(0) = 5(0)2 - 10
5 = 0 - 10
5 ≠ -10
So, the given ordered pair (0, 5) is not a solution.
Problem 4 :
f(x) = 2x2 + 5x - 3 (4, 9)
Solution :
f(x) = 2x2 + 5x - 3
Given ordered pair is (4, 9)
f(4) = 2(4)2 + 5(4) - 3
9 = 2(16) + 20 - 3
9 = 32 + 20 - 3
9 = 52 - 3
9 ≠ 49
So, the given ordered pair (4, 9) is not a solution.
Problem 5 :
f(x) = -2x2 + 3x (-1/2, 1)
Solution :
f(x) = -2x2 + 3x
Given ordered pair is (-1/2, 1)
f(-1/2) = -2(-1/2)2 + 3(-1/2)
1 = -2(1/4) - (3/2)
1 = -1/2 - (3/2)
1 = (-1-3)/2
1 = -4/2
1 ≠ -2
So, the given ordered pair (-1/2, 1) is not a solution.
Problem 6 :
f(x) = -7x2 + 8x + 15 (-1, 16)
Solution :
f(x) = -7x2 + 8x + 15
Given ordered pair is (-1, 16)
f(-1) = -7(-1)2 + 8(-1) + 15
16 = -7(1) - 8 + 15
16 = -7 - 8 + 15
16 = -15 + 15
16 ≠ 0
So, (-1, 16) is not a solution.
Problem 7 :
f(x) = 3x2 - 13x + 4 (2, -10)
Solution :
f(x) = 3x2 - 13x + 4
Given ordered pair is (2, -10)
f(2) = 3(2)2 - 13(2) + 4
-10 = 3(4) - 26 + 4
-10 = 12 - 26 + 4
-10 = 16 - 26
-10 = -10
(2, -10) is a solution.
Find the values of x for the given value of y for each of the following quadratic function.
Problem 8 :
y = x2 + 6x + 10 {y = 1}
Solution :
y = x2 + 6x + 10
Applying y = 1, we get
1 = x2 + 6x + 10
x2 + 6x + 10 - 1 = 0
x2 + 6x + 9 = 0
x2 + 3x + 3x + 9 = 0
x(x + 3) + 3(x + 3) = 0
(x + 3)(x + 3) = 0
x = -3 and x = -3
The solutions are (-3, 1) and (-3, 1)
Problem 9 :
y = x2 + 5x + 8 {y = 2}
Solution :
y = x2 + 5x + 8
Applying y = 2, we get
2 = x2 + 5x + 8
x2 + 5x + 8 - 2 = 0
x2 + 5x + 6 = 0
x2 + 2x + 3x + 6 = 0
x(x + 2) + 3(x + 2) = 0
(x + 2)(x + 3) = 0
x = -2 and x = -3
The solutions are (-2, 2) and (-3, 2)
Problem 10 :
y = x2 - 5x + 1 {y = -3}
Solution :
y = x2 - 5x + 1
Applying y = -3, we get
-3 = x2 - 5x + 1
x2 - 5x + 1 + 3 = 0
x2 - 5x + 4 = 0
x2 - 1x - 4x + 4 = 0
x(x - 1) - 4(x - 1) = 0
(x - 1)(x - 4) = 0
x = 1 and x = 1
The solutions are (1, -3) and (1, -3)
Problem 11 :
f(x) = x2 + 4x + 11 find x when
i) f(x) = 23 ii) f(x) = 7
Solution :
f(x) = x2 + 4x + 11
i) f(x) = 23
23 = x2 + 4x + 11
x2 + 4x + 11 - 23 = 0
x2 + 4x - 12 = 0
x2 + 6x - 2x - 12 = 0
x(x + 6) - 2(x + 6) = 0
(x + 6)(x - 2) = 0
x = -6 and x = 2
So, the solutions are (-6, 23) and (2, 23).
ii) f(x) = 7
7 = x2 + 4x + 11
x2 + 4x + 11 - 7 = 0
x2 + 4x + 4 = 0
x2 + 2x + 2x + 4 = 0
x(x + 2) + 2(x + 2) = 0
(x + 2)(x + 2) = 0
x = -2 and x = -2
So, the solutions are (-2, 7) and (-2, 7).
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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