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Oxana [17]
2 years ago
12

Given that a and b are the zeros of the polynomial f(x) = x^2 - x - 6 with a > b, and that g(x) = f(x + 2) find:

Mathematics
1 answer:
artcher [175]2 years ago
8 0

Answer:he end behavior of a function fff describes the behavior of its graph at the "ends" of the xxx-axis. Algebraically, end behavior is determined by the following two questions:

As x\rightarrow +\inftyx→+∞x, right arrow, plus, infinity, what does f(x)f(x)f, left parenthesis, x, right parenthesis approach?

As x\rightarrow -\inftyx→−∞x, right arrow, minus, infinity, what does f(x)f(x)f, left parenthesis, x, right parenthesis approach?

If this is new to you, we recommend that you check out our end behavior of polynomials article.

The zeros of a function fff correspond to the xxx-intercepts of its graph. If fff has a zero of odd multiplicity, its graph will cross the xxx-axis at that xxx value. If fff has a zero of even multiplicity, its graph will touch the xxx-axis at that point.

If this is new to you, we recommend that you check out our zeros of polynomials article.

Step-by-step explanation:. PA BRAINLIEST

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2 years ago
Analyze a graph In Exercise,analyze and sketch the graph of the function.Label any relative extrema, points of inflation,and asy
ZanzabumX [31]

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Step-by-step explanation:

Given is a function f(x)

f(x) = e^x -e^{-\frac{{x}}{2}

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Y intercept: Put x =0 , we get y =0

No exception for x or y and hence set of all real numbers is domain and range.

f'(x) = (e^x +1/2 e^{-\frac{{x}}{2})

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8 0
3 years ago
What is 21 > 15 + 2a
raketka [301]
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5 0
3 years ago
Simplify the rational expression. State any restrictions on the variable.
ExtremeBDS [4]

we are given

\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}

Firstly, we will factor numerators and denominators

\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-6)(n^2-4)}{(n^2-3)(n^2-6)}

we can see that

n^2 -6 is factor on both numerator and denominator

so, it will get cancelled

and n^2 -6 can not be equal to 0

so, one of restriction is

n^2-6\neq 0

n\neq -+ \sqrt{6}

we can simplify it

\frac{(n^4-10n^2+24)}{(n^4-9n^2+18)}=\frac{(n^2-4)}{(n^2-3)}

we know that denominator can not be zero

n^2-3\neq 0

n\neq -+ \sqrt{3}

so, option-B.......Answer

3 0
3 years ago
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