hope u get it. I've used RHS to prove LHS
9 is your answer, hope this helps
9 - 6 + 4 - 8/3 ..,
geometric series a(n) = a1r^(n-1)
r = a(n+1)/a(n)
-6/9 = -2/3
4/-6 = -2/3
-8/3/4 = -2/3
so r = -2/3 and a1 = 9
Sn = a1(1-r^n)/(1-r) = 9(1-(-2/3)^n)/(1-(-2/3))
n is infinite Sn = 9/(5/3) = 27/5
Answer: the graph crosses the x-axis at x = -3
<u>Step-by-step explanation:</u>
y = (x + 3)³
To find where the graph crosses the x-axis, let y = 0 and solve for x:
0 = (x + 3)³
0 = (x + 3) with a multiplicity of 3
-3 = x with a multiplicity of 3.
Since multiplicity is an ODD number, the graph CROSSES the x-axis at x = -3
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<u>Graph:</u>
- Leading coefficient is POSITIVE so right side goes to +∞
- Degree of polynomial is ODD so left side goes to -∞
<em>graph is attached</em>