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
F(x)=x-5
g(x)=x^2
Now adding f(x) and g(x)
x-5+x^2-1
x^2+x-6
x^2+(3-2)x-6
x^2+3x-2x-6
x(x+3)-2(x+3)
(x-2)(x+3) answer
4(60q-190u-13b+17i) you have to factor the polynomials
The answer to the question is 3x+97
Let 
. Then
lies in the second quadrant, so
So we have
and the fourth roots of are
where 
. In particular, they are
