Find a polynomial function whose graph passes through (6,13), (9,- 12), and (0,4)
Well....we can generate a quadratic for this
Let the polynomial be
y = ax^2 + bx + c
Since (0, 4) is in the graph, then c = 4
And using the other two points we have this system
13 = a(6)^2 + b(6) + 4
-12 = a(9)^2 + b(9) + 4 simplify
36a + 6b = 9
81a + 9b = - 16
Divide the first equation through by 6
6a + b = 3/2
b = (3/2) - 6a sub this into the second equation for b and we have that
81a +9(3/2 -6a) = -16
81a + 27/2 - 54a = -16
27a = -16 - 27/2
27a = [ -32 - 27 ] / 2
27a = [ -59 ] /2
a = [ -59/27] / 2
a = -59/54
And
b = (3/2) - 6 (-59/54 )
b = (3/2) + 59/9
b = 145/18
So.....the polynomial is
-(59/54)x^2 + (145/18)x + 4
Here's a graph : https://www.desmos.com/calculator/jrc6i4llx4