Exercise 1.2.11 (11-12)

Find a formula for the quadratic function whose graph is shown.

Answers

11.
Since f is a quadratic function, we can find coefficients a , b , c ∈ ℝ such that ∀ ⁡ x ∈ ℝ : f ( x ) = a x 2 + bx + c .

Since the points ( 0 , 18 ) , ( 3 , 0 ) , ( 4 , 2 ) lie on the graph of f , we get the following three equations:

f ( 0 ) = 18 ⟹ a ⋅ 0 2 + b ⋅ 0 + c = 18 . 		(1)

f ( 3 ) = 0 ⟹ a ⋅ 3 2 + b ⋅ 3 + c = 0 . 		(2)

f ( 4 ) = 2 ⟹ a ⋅ 4 2 + b ⋅ 4 + c = 2 . 		(3)

From equation 1 , we have:

0 + 0 + c = 18 .

c = 18 . 		(4)

From the equation 2 we have:

9 a + 3 b + c = 0 .

Since from 4 we have c = 18 , we can substitute this value in the previous equation:

9 a + 3 b + 18 = 0 .

9 a + 3 b = − 18 . ( ÷ 3 )

3 a + b = − 6 . 		(5)

From the equation 3 we have:

16 a + 4 b + c = 2 .

Since from 4 we have c = 18 , we have:

16 a + 4 b + 18 = 2 .

16 a + 4 b = − 16 . ( ÷ 4 )

4 a + b = − 4 . 		(6)

To find the values of a and b we should solve the system of the equations 5 and 6 :

{ 3 a + b = − 6 4 a + b = − 4 ⟺ { 3 a + b − ( 4 a + b ) = − 6 − ( − 4 ) 4 a + b = − 4 ⟺ { − a = − 2 4 a + b = − 4 ⟺

⟺ { a = 2 4 ⋅ 2 + b = − 4 ⟺ { a = 2 b = − 12 .

Thus, the graph in Picture 11 describes the function f ( x ) = 2 x 2 − 12 x + 18 .

12.
We can solve (12) using the same logic as in (11). Since f is a quadratic function, we can find coefficients a , b , c ∈ ℝ such that ∀ ⁡ x ∈ ℝ : f ( x ) = a x 2 + bx + c .

Since the points ( − 2 , 2 ) , ( 0 , 1 ) , ( 1 , − 2.5 ) lie on the graph of f , we get the following three equations:

f ( − 2 ) = 2 ⟹ a ⋅ ( − 2 ) 2 + b ⋅ ( − 2 ) + c = 2 . 		(7)

f ( 0 ) = 1 ⟹ a ⋅ 0 2 + b ⋅ 0 + c = 1 . 		(8)

f ( 1 ) = − 2.5 ⟹ a ⋅ 1 2 + b ⋅ 1 + c = − 2.5 . 		(9)

From equation 8 , we have:

0 + 0 + c = 1 .

c = 1 . 		(10)

From the equation 7 we have:

4 a − 2 b + c = 2 .

Since from 10 we have c = 1 , we can substitute this value to the previous equation:

4 a − 2 b + 1 = 2 .

4 a − 2 b = 1 . 		(11)

From the equation 9 we have:

a + b + c = − 2.5 .

Since from 10 we have c = 18 , we have:

a + b + 1 = − 2.5 .

a + b = − 3.5 . 		(12)

To find the values of a and b we should solve the system of the equations 11 and 12 :

{ a + b = − 3.5 4 a − 2 b = 1 ⟺ { a + b = − 3.5 2 a − b = 0.5 ⟺ { 3 a = − 3 a + b = − 3.5 ⟺

⟺ { a = − 1 b = − 2.5

Thus, the graph in Picture 12 describes the function f ( x ) = − x 2 − 2.5 x + 1 .

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2023-08-05 06:53
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