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Alenkinab [10]
3 years ago
7

Determine f’s end behavior f(x)= -5x^6+8x^5-1/ x^2 -9

Mathematics
1 answer:
kati45 [8]3 years ago
4 0

Answer:

f(x) → -∞ as x → -∞

f(x) → -∞ as x → +∞

Step-by-step explanation:

The given function is;

f(x) = -5x^(6) + 8x^(5) - 1/(x² - 9x)

Using long division to divide this as attached, we have;

f(x) = -5x⁴ - 37x³ - 333x² - 2997x - 26973 + (-242757x - 1)/(x² - 9x)

Thus, the leading coefficient here is -5 and the polynomial degree is 4.

Since the leading coefficient is negative and the degree of the polynomial is an even number, then we can say that the behavior of the polynomial is;

f(x) → -∞ as x → -∞

f(x) → -∞ as x → +∞

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Find the simplified product b-5/2b x b^2+3b/b-5
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Answer:

The product \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}

Step-by-step explanation:

Given expression \frac{b-5}{2b} and \frac{b^2+3b}{b-5}

We have to find the product of  \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}

   

Consider the given expression  \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}

Multiply fractions, we have,

\frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d}

=\frac{\left(b-5\right)\left(b^2+3b\right)}{2b\left(b-5\right)}

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we have, =\frac{b^2+3b}{2b}

Apply exponent rule,

\:a^{b+c}=a^ba^c

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=\frac{b\left(b+3\right)}{2b}

Cancel common factor b , we have,

=\frac{b+3}{2}

Thus, the product  \frac{b-5}{2b}\times\frac{b^2+3b}{b-5}=\frac{b+3}{2}

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3 years ago
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Using compatible numbers 911-347
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Hi, 911 - 347 = 564.

Hope this helps! :)
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