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Answer:
Find a polynomial function whose graph passes through (6,13), (9,-11), (0,5)
1 Answers
Assuming a quadratic, we have that
y = ax^2 + bx + c
Since (0,5) is on the graph, c =5
And we have the remaining system
a(9)^2 + b(9) + 5 = -11
a(6)^2 + b(6) + 5 = 13 simplify
81a + 9b = -16 multiply through by 6 ⇒ 486a + 54b = - 96 (1)
36a + 6b = 8 multiply through by -9 ⇒ -324a -54b = -72 (2)
Add (1) and (2)
162a = -168
a = -28/27
To find b we have
36 (-28/27) + 6b = 8
-112/3 + 6b = 8
⇒ b = 68/9
The function is
y = - (28/27)x^2 + (68/9)x + 5
Answer:
See below
Step-by-step explanation:
Let the function be quadratic as the given points are not collinear, not a linear function
Standard form of quadratic equation:
- y = ax^2 + bx + c
Substituting the coordinates, we get:
- 13 = a(6)^2 + b(6) + c
- -10 = a(10)^2 + b(10) + c
- 5 = a(0)^2 + b(0) + c
From equation 3 we get c = 5, considering this in the other equations and simplifying:
- 13 = 36a + 6b + 5 ⇒ 36a + 6b = 8 ⇒ 18a + 3b = 4
- -10 = 100a + 10b + 5 ⇒ 100a + 10b = -15 ⇒ 20a + 2b = -3
Subtract 3 times the second equation from 2 times the first:
- 2(18a) + 2(3b) - 3(20a) - 3(2b) = 4(2) - 3(-3)
- 36a - 60a = 8 + 9
- -24a = 17
- a = - 17/24
Finding the value of b:
- 18(-17/24) + 3b = 4
- 3b = 4 + 51/4
- b = 67/12
The quadratic function is:
- y = -17/24x^2 + 67/12x + 5
So the graph of any polynomial function:
- y = -17/24x^2 + 67/12x + 5 + g(x)(x-6)(x-10)(x-0)
will pass through the given 3 points