Zeros are the x-values that make the function zero.
0 = (x-0)(x- -4)(x-3)
f(x) = x(x+4)(x-3)
Answer:
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Step-by-step explanation:
Answer:
A.
Step-by-step explanation:
if to re-write all the equations:
A. y=3/2 x-3; y=3/2 x-3; B. y=-7x+10; y=3/2 x+3/2; C. y=x+5; y=-3/4 x+2; D. y=x+1; y= -3/4 x+8.
According to the re-written view the correct answer is 'A'.
Let

, so that


Substituting into the ODE gives


The first series starts with a linear term, while the other two start with a constant. Extract the first term from each of the latter two series:


Finally, to get the series to start at the same index, shift the index of the first two series by replacing

with

. Then the ODE becomes

which can be consolidated to get
![\displaystyle\sum_{k\ge1}\bigg[(3k(k+1)+6(k+1))a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bk%5Cge1%7D%5Cbigg%5B%283k%28k%2B1%29%2B6%28k%2B1%29%29a_%7Bk%2B1%7D%2Ba_k%5Cbigg%5Dx%5Ek%2B6a_1%2Ba_0%3D0)
![\displaystyle\sum_{k\ge1}\bigg[3(k+1)(k+2)a_{k+1}+a_k\bigg]x^k+6a_1+a_0=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bk%5Cge1%7D%5Cbigg%5B3%28k%2B1%29%28k%2B2%29a_%7Bk%2B1%7D%2Ba_k%5Cbigg%5Dx%5Ek%2B6a_1%2Ba_0%3D0)
You're fixing the solution so that it contains the origin, which means

which in turn means

. With the given recurrence, it follows that

for all

, so the solution would be

. This is to be expected, since

is clearly a singular point for the ODE.