What transformation(s) were made to the original f(x) = x graph?
1 answer:
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Answer:
A, B, and E
if I read your functions right.
Step-by-step explanation:
It's zeros are x=-6,-2, and 2.
This means we want the factors (x+6) and (x+2) and (x-2) in the numerator.
It has a y-intercept of 4. This means we want to get 4 when we plug in 0 for x.
And it's long-run behavior is y approaches - infinity as x approaches either infinity. This means the degree will be even and the coefficient of the leading term needs to be negative.
So let's see which functions qualify:
A) The degree is 4 because when you do x^2*x*x you get x^4.
The leading coefficient is -4/144 which is negative.
We do have the factors (x+6), (x+2), and (x-2).
What do we get when plug in 0 for x:
Put into calculator: 4
A works!
B) The degree is 6 because when you do x*x^4*x=x^6.
The leading coefficient is -4/192 which is negative.
We do have factors (x+6), (x+2), and (x-2).
What do we get when we plug in 0 for x:
Put into calculator: 4
B works!
C) The degree is 4 because when you do x*x*x*x=x^4.
The leading coefficient is -4 which is negative.
Oops! It has a zero at 0 because of that factor of (x) between -4 and (x+6).
So C doesn't work.
D) The degree is 3 because x*x*x=x^3.
We needed an even degree.
D doesn't work.
E) The degree is 4 because x*x^2*x=x^4.
The leading coefficient is -4/48 which is negative.
It does have the factors (x+6), (x+2), and (x-2).
What do we get when we plug in 0 for x:
Put into calculator: 4
So E does work.
F) The degree is 4 because x*x*x^2=x^4.
The leading coefficient is -4/48.
It does have factors (x+6), (x+2), and (x-2).
What do we get when we plug in 0 for x:
Put into calculator: -4
So F doesn't work.
G. I'm not going to go any further. The leading coefficient is 4/48 and that is not negative.
So G doesn't work.