If u add up 15,17, and 19 they add up to 51
I think is inside and then outside. Not sure
Hope this helps!
Answer:
C 2
Step-by-step explanation:
This is a parabolic function
Notice how we go negative and then positive
y = ax^2 +bx+c
Let x = 0
-5 = a(0) + b(0) +c
c = -5
y = ax^2 + bx -5
Let x = 3
4 = a(3)^2 +b(3) -5
Let x=-3
4 = a(3)^2 -b(3) -5
Add the two equations
4 = a(3)^2 +b(3) -5
4 = a(3)^2 -b(3) -5
----------------------------
8 = 2a (3)^2 - 10
18 = 2 a(9)
18 = 18a
a =1
Solving for b
4 = 1(3)^2 -b(3) -5
4 = 9 -3b -5
4 = 4 -3b
0 = -3b
0 =b
The equation is
y = (x)^2 -5
Letting y = -1
-1 =x^2 -5
Adding 5 to each side
4 = x^2
Taking the square root of each side
±2 = x
x= ±2
Given the choices
For zeroes of r1,r2,r3, the factors of the function, f(x) are
(x-r1)(x-r2)(x-r3)
zeroes of -3,-5,2
(x-(-3))(x-(-5))(x-2)=
(x+3)(x+5)(x-2)
f(x)=(x+3)(x+5)(x-2)
expanded
f(x)=x³+6x²-x-30
Answer:
<h3>a. Yes, they are part of the solution</h3>
