Question

Find a polynomial function whose graph passes through ​(6,13),(10,-10) and (0,5)

Answer 1

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:

1. 13 = a(6)^2 + b(6) + c
2. -10 = a(10)^2 + b(10) + c
3. 5 = a(0)^2 + b(0) + c

From equation 3 we get c = 5, considering this in the other equations and simplifying:

1. 13 = 36a + 6b + 5 ⇒ 36a + 6b = 8 ⇒ 18a + 3b = 4
2. -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

Answer 2

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