Question

algebra 2 unit 3: end-of-unit assessment (1a)
graphing calculators should not be used. use of a four-function or scientific calculator is
acceptable. show all your work to receive full credit!!!
i can find equivalent expressions and use them to discover end behavior of an expression

1. select the expression that is equivalent to \\(\frac{2x^3 + 11x^2 - 21x}{x^2 + 3x}\\) for \\(x \

eq 3\\) or 0. 1 point____
a. \\(2x + 5 - \frac{6}{x + 3}\\)
b. \\(2x + 17 - \frac{20}{x + 3}\\)
c. \\(2x + 17 - \frac{36}{x^2 + 3x}\\)
d. \\(2x + 5 - \frac{36}{x + 3}\\)

1. what is the end behavior of the above expression?

1 point____
© 2024 illustrative mathematics. includes additional content © by imagine learning

Explanation:

Question 1

Step 1: Factor Numerator and Denominator

First, factor the numerator \(2x^3 + 11x^2 - 21x\) and the denominator \(x^2 + 3x\).

• Numerator: \(2x^3 + 11x^2 - 21x = x(2x^2 + 11x - 21)\). Factor \(2x^2 + 11x - 21\): find two numbers that multiply to \(2(-21)=-42\) and add to \(11\). These numbers are \(14\) and \(-3\). So, \(2x^2 + 14x - 3x - 21 = 2x(x + 7) - 3(x + 7) = (2x - 3)(x + 7)\). Wait, maybe polynomial long division is better.

Denominator: \(x^2 + 3x = x(x + 3)\).

Step 2: Perform Polynomial Long Division

Divide \(2x^3 + 11x^2 - 21x\) by \(x^2 + 3x\).

• Divide \(2x^3\) by \(x^2\) to get \(2x\). Multiply \(2x\) by \(x^2 + 3x\): \(2x(x^2 + 3x) = 2x^3 + 6x^2\). Subtract from the numerator: \((2x^3 + 11x^2 - 21x) - (2x^3 + 6x^2) = 5x^2 - 21x\).
• Now divide \(5x^2\) by \(x^2\) to get \(5\). Multiply \(5\) by \(x^2 + 3x\): \(5(x^2 + 3x) = 5x^2 + 15x\). Subtract: \((5x^2 - 21x) - (5x^2 + 15x) = -36x\).
• Now we have a remainder of \(-36x\) and the divisor is \(x^2 + 3x\). Wait, but let's factor the denominator again. Wait, maybe factor numerator and denominator:

Numerator: \(x(2x^2 + 11x - 21) = x(2x - 3)(x + 7)\) (wait, no, earlier factoring was wrong). Wait, let's try again: \(2x^2 + 11x - 21\). Using quadratic formula: \(x = \frac{-11 \pm \sqrt{121 + 168}}{4} = \frac{-11 \pm \sqrt{289}}{4} = \frac{-11 \pm 17}{4}\). So \(x = \frac{6}{4} = \frac{3}{2}\) or \(x = -7\). So \(2x^2 + 11x - 21 = (2x - 3)(x + 7)\). Denominator: \(x(x + 3)\). Hmm, maybe long division is better.

Wait, let's do long division properly:
Dividend: \(2x^3 + 11x^2 - 21x\) (degree 3)
Divisor: \(x^2 + 3x\) (degree 2)

• First term: \(2x^3 / x^2 = 2x\). Multiply divisor by \(2x\): \(2x \cdot x^2 + 2x \cdot 3x = 2x^3 + 6x^2\). Subtract from dividend:

\((2x^3 + 11x^2 - 21x) - (2x^3 + 6x^2) = 5x^2 - 21x\).

• Next term: \(5x^2 / x^2 = 5\). Multiply divisor by \(5\): \(5x^2 + 15x\). Subtract:

\((5x^2 - 21x) - (5x^2 + 15x) = -36x\).

Now, the division is \(2x + 5 + \frac{-36x}{x^2 + 3x}\). Simplify the remainder: \(\frac{-36x}{x(x + 3)} = \frac{-36}{x + 3}\) (since \(x
eq 0\)). So the expression becomes \(2x + 5 - \frac{36}{x + 3}\)? Wait, no, wait: the remainder is \(-36x\), divisor is \(x^2 + 3x = x(x + 3)\), so \(\frac{-36x}{x(x + 3)} = \frac{-36}{x + 3}\). Wait, but let's check the options. Option D is \(2x + 5 - \frac{36}{x + 3}\)? Wait, no, option D is \(2x + 5 - \frac{36}{x + 3}\)? Wait, the options are:
A. \(2x + 5 - \frac{6}{x + 3}\)
B. \(2x + 17 - \frac{20}{x + 3}\)
C. \(2x + 17 - \frac{36}{x^2 + 3x}\)
D. \(2x + 5 - \frac{36}{x + 3}\)

Wait, maybe I made a mistake in long division. Let's try factoring numerator and denominator:

Numerator: \(2x^3 + 11x^2 - 21x = x(2x^2 + 11x - 21)\)
Denominator: \(x^2 + 3x = x(x + 3)\)

Cancel \(x\) (since \(x
eq 0\)): \(\frac{2x^2 + 11x - 21}{x + 3}\)

Now perform long division on \(\frac{2x^2 + 11x - 21}{x + 3}\):

• Divide \(2x^2\) by \(x\) to get \(2x\). Multiply \(2x\) by \(x + 3\): \(2x^2 + 6x\). Subtract: \((2x^2 + 11x - 21) - (2x^2 + 6x) = 5x - 21\).
• Divide \(5x\) by \(x\) to get \(5\). Multiply \(5\) by \(x + 3\): \(5x + 15\). Subtract: \((5x - 21) - (5x + 15) = -36\).

So the division gives \(2x + 5\) with a remainder of \(-36\), so \(\frac{2x^2 + 11x - 21}{x + 3} = 2x + 5 - \frac{36}{x + 3}\). Since we canceled \(x\) earlier (from numerator and denominator), the original expression is \(2x + 5 - \frac{36}{x + 3}\) (since \(x
eq 0\) or \(3\)). So the correct option is D.

To determine the end behavior of a rational function, we analyze the degrees of the numerator and denominator.

• The original expression is \(\frac{2x^3 + 11x^2 - 21x}{x^2 + 3x}\). Simplify by dividing numerator and denominator by \(x^2\) (for \(x

eq 0\)):
\(\frac{2x + 11 - \frac{21}{x}}{1 + \frac{3}{x}}\).

As \(x \to \infty\) or \(x \to -\infty\), the terms with \(\frac{1}{x}\) (i.e., \(-\frac{21}{x}\) and \(\frac{3}{x}\)) approach \(0\). So the expression behaves like \(\frac{2x + 11}{1} = 2x + 11\) as \(x \to \pm\infty\). Wait, but from the equivalent expression in part 1, \(2x + 5 - \frac{36}{x + 3}\), as \(x \to \infty\) or \(x \to -\infty\), the term \(-\frac{36}{x + 3}\) approaches \(0\), so the end behavior is dominated by \(2x + 5\) (which is linear, same as \(2x + 11\) in the simplified form, but the leading term is \(2x\)).

For a rational function \(\frac{f(x)}{g(x)}\) where \(\text{deg}(f) = n\) and \(\text{deg}(g) = m\):

• If \(n > m\), the end behavior is like the quotient of the leading terms (since the rational function behaves like a polynomial of degree \(n - m\)). Here, numerator degree is \(3\), denominator degree is \(2\), so \(n - m = 1\). The leading term of the numerator is \(2x^3\), leading term of the denominator is \(x^2\), so the quotient of leading terms is \(\frac{2x^3}{x^2} = 2x\). Thus, as \(x \to \infty\), \(y \to \infty\); as \(x \to -\infty\), \(y \to -\infty\) (since the leading coefficient \(2\) is positive, and the degree of the quotient polynomial is \(1\) (odd), so the end behavior is \(x \to \infty\), \(f(x) \to \infty\); \(x \to -\infty\), \(f(x) \to -\infty\)).

Answer:

D. \(2x + 5 - \frac{36}{x + 3}\)

Question 2