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

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) + c5 = 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 = 17a = - 17/24

Finding the value of b:

18(-17/24) + 3b = 43b = 4 + 51/4b = 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

hope it is correct if not i'm srry i tried to

Ujjg