Zeros of Polynomials to Construct Graphs - Algebra 2

The polynomial $f(x) = 2x^3 - 8x^2 + 6x$ can be factored completely. Which statement best describes the behavior of the graph of $y = f(x)$ near its zeros?
Answer: The graph crosses the x-axis at all three zeros: $x = 0$ , $x = 1$ , and $x = 3$ First factor: $f(x) = 2x^3 - 8x^2 + 6x = 2x(x^2 - 4x + 3) = 2x(x-1)(x-3)$ . The zeros are $x = 0, 1, 3$ , each with multiplicity 1 (odd). Since all multiplicities are odd, the graph crosses the x-axis at each zero. Choice A incorrectly suggests $x = 1$ has even multiplicity. Choice C incorrectly suggests $x = 0$ has even multiplicity. Choice D incorrectly suggests $x = 3$ has even multiplicity.
A cubic polynomial $s(x)$ has exactly two x-intercepts visible on its graph, with the graph crossing the x-axis at $x = -1$ and appearing to just touch the x-axis at $x = 3$ . Which factored form best represents this polynomial?
Answer: $s(x) = a(x+1)(x-3)^2$ where $a$ is a nonzero constant When analyzing polynomial graphs, the key is understanding how the behavior at each x-intercept reveals information about the multiplicity of that zero. If the graph crosses the x-axis, the zero has odd multiplicity. If it touches but doesn't cross (creating a "bounce"), the zero has even multiplicity. Since the graph crosses at $x = -1$ , this zero has odd multiplicity. Since it touches but doesn't cross at $x = 3$ , this zero has even multiplicity. For a cubic polynomial with only two visible x-intercepts, you need the multiplicities to add up to 3. The only way to achieve this is with multiplicities of 1 and 2. Since $x = -1$ crosses (odd), it has multiplicity 1. Since $x = 3$ touches (even), it has multiplicity 2. This gives us $s(x) = a(x+1)(x-3)^2$ , which is answer choice D. Choice A is incorrect because it only has one real zero at $x = -1$ with multiplicity 3, not two visible x-intercepts. Choice B reverses the multiplicities—it would show touching at $x = -1$ and crossing at $x = 3$ , opposite of what's described. Choice C suggests three distinct real zeros, which would produce three x-intercepts, not two. Remember this pattern: crossing the x-axis means odd multiplicity, while touching (bouncing off) means even multiplicity. For cubic polynomials, always check that the sum of multiplicities equals 3.
Describe the graph's behavior based on factored form for $p(x)=(x+2)(x-1)^2(x-3).$ Which statement is correct about zeros, multiplicities, crossing/touching, and end behavior?
Answer: Zeros: $x=-2$ (mult. 1), $x=1$ (mult. 2), $x=3$ (mult. 1). Crosses at $-2$ and $3$ , touches at $1$ . End behavior: as $x \to -\infty$ , $p(x) \to \infty$ and as $x \to \infty$ , $p(x) \to \infty$ . This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form $p(x) = a(x - r_1)(x - r_2) \dots$ are found by setting each factor equal to zero: from $(x - r)$ , the zero is $x = r$ . Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For $p(x) = (x+2)(x-1)^2(x-3)$ , we have zeros at $x = -2$ (multiplicity 1, from $(x+2)$ ), $x = 1$ (multiplicity 2, from $(x-1)^2$ ), and $x = 3$ (multiplicity 1, from $(x-3)$ ). The graph crosses at $x = -2$ and $x = 3$ (odd multiplicities) and touches at $x = 1$ (even multiplicity). The polynomial has degree $1+2+1=4$ (even) with positive leading coefficient, so both ends go up: as $x \to -\infty$ , $p(x) \to \infty$ and as $x \to \infty$ , $p(x) \to \infty$ . Choice A correctly identifies all zeros with their multiplicities, states that the graph crosses at $-2$ and $3$ while touching at $1$ , and has the correct end behavior for an even-degree polynomial with positive leading coefficient. Choice C incorrectly claims the graph touches at all zeros, which would require all multiplicities to be even. Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero.
For $g(x)=x^2(x-5)^3,$ identify the zeros and their multiplicities, describe how the graph behaves at each zero (crosses or touches, and whether it flattens), and state the end behavior.
Answer: Zeros: $x=0$ (mult. 2) and $x=5$ (mult. 3). Touches/bounces at $0$ ; crosses with flattening at $5$ . End behavior: left down, right up. This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For g(x) = $x^2$ $(x - 5)^3$ , the zeros are x = 0 (multiplicity 2, from $x^2$ = 0) and x = 5 (multiplicity 3, from $(x - 5)^3$ = 0); the graph touches and bounces at x = 0 due to even multiplicity and crosses with flattening at x = 5 due to odd multiplicity greater than 1, with the degree 5 (odd) and positive leading coefficient indicating left down and right up. Choice A correctly identifies the zeros with multiplicities, shows proper touching and crossing with flattening, and has correct end behavior. Choice C fails because it swaps the multiplicities, stating x = 0 has multiplicity 3 and x = 5 has 2, which reverses the behaviors at those zeros. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!
For $p(x)=-(x+4)^2(x-1)^3$ , sketch $p(x)$ showing zeros, multiplicities, and end behavior. Which statement is correct?
Answer: Zeros: $x=-4$ (mult. 2), $x=1$ (mult. 3). Touches at $x=-4$ , crosses (flattened) at $x=1$ . End behavior: as $x\to -\infty$ , $p(x)\to +\infty$ ; as $x\to +\infty$ , $p(x)\to -\infty$ . This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. For p(x) = -(x+4)²(x-1)³, we find zeros: (x+4)²=0 gives x=-4 (multiplicity 2), and (x-1)³=0 gives x=1 (multiplicity 3). At x=-4, multiplicity 2 is even, so the graph touches and bounces. At x=1, multiplicity 3 is odd, so the graph crosses with flattening. The polynomial has degree 2+3=5 (odd) with negative leading coefficient (-1), so as x→-∞, p(x)→+∞ (left up) and as x→+∞, p(x)→-∞ (right down). Choice A correctly identifies zeros x=-4 (mult. 2), x=1 (mult. 3), shows touching at x=-4 and crossing (flattened) at x=1, with end behavior as x→-∞, p(x)→+∞ and as x→+∞, p(x)→-∞. Choice D incorrectly identifies x=4 instead of x=-4 as a zero, missing the sign in the factor (x+4). The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!
Sketch $f(x)=-2(x+3)^2(x-1)(x-4)$ showing zeros, their multiplicities, and end behavior. Which description matches the correct rough graph?
Answer: Zeros: $x=-3$ (mult. 2), $x=1$ (mult. 1), $x=4$ (mult. 1). Touches at $-3$ , crosses at $1$ and $4$ . End behavior: as $x\to -\infty$ , $f(x)\to -\infty$ and as $x\to +\infty$ , $f(x)\to -\infty$ . This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For f(x) = -2 $(x + 3)^2$ (x - 1)(x - 4), zeros are x = -3 (multiplicity 2, touches), x = 1 (multiplicity 1, crosses), x = 4 (multiplicity 1, crosses), with degree 4 (even) and negative leading coefficient, so both ends down. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. A distractor like choice B might get the end behavior wrong by assuming odd degree, but count the total degree as 4 (even) and note the negative coefficient flips it downward—keep practicing to spot this! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!
Use zeros to construct a rough graph of $p(x)=-x(x+2)^2(x-5).$ Identify the zeros with multiplicities, determine crossing vs touching, and state the end behavior.
Answer: Zeros: $x=0$ (mult. 1), $x=-2$ (mult. 2), $x=5$ (mult. 1). Crosses at $0$ and $5$ , touches at $-2$ . End behavior: as $x\to -\infty$ , $p(x)\to -\infty$ and as $x\to +\infty$ , $p(x)\to -\infty$ . This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For p(x) = - $x(x + 2)^2$ (x - 5), zeros are x = 0 (multiplicity 1, crosses), x = -2 (multiplicity 2, touches), x = 5 (multiplicity 1, crosses), degree 4 even negative, both ends down. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice B assumes crossing at all but misses that even multiplicity touches—count exponents carefully to avoid this! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!
Sketch $f(x)=(x-3)(x+1)(x-2)^2$ showing zeros, multiplicities, and end behavior. Which option matches the correct sketch description?
Answer: Zeros: $x=3$ (mult. 1), $x=-1$ (mult. 1), $x=2$ (mult. 2). Crosses at $x=3$ and $x=-1$ , touches at $x=2$ . End behavior: as $x\to -\infty$ , $f(x)\to +\infty$ ; as $x\to +\infty$ , $f(x)\to +\infty$ . This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. For f(x) = (x-3)(x+1)(x-2)², we find zeros: x-3=0 gives x=3 (multiplicity 1), x+1=0 gives x=-1 (multiplicity 1), and (x-2)²=0 gives x=2 (multiplicity 2). The graph crosses at x=3 and x=-1 (odd multiplicities) and touches at x=2 (even multiplicity). The polynomial has degree 1+1+2=4 (even) with positive leading coefficient (1), so both ends go up: as x→±∞, f(x)→+∞. Choice A correctly identifies zeros x=3 (mult. 1), x=-1 (mult. 1), x=2 (mult. 2), shows crossing at x=3 and x=-1, touching at x=2, and has both ends going to +∞. Choice B incorrectly claims the graph crosses at all three zeros, missing that x=2 with even multiplicity should touch, not cross. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!
If $r(x) = x^3 - 7x^2 + 14x - 8$ and it is known that $x = 1$ is a zero, which statement correctly describes the complete factorization and the resulting graph behavior?
Answer: $r(x) = (x-1)(x-2)(x-4)$ , and the graph crosses the x-axis at three distinct points When you encounter a polynomial with a known zero, your task is to use polynomial division or synthetic division to find the remaining factors, then analyze how the graph behaves at each zero based on the multiplicity of those factors. Since $x = 1$ is a zero, $(x-1)$ is a factor. Using synthetic division or polynomial long division to divide $r(x) = x^3 - 7x^2 + 14x - 8$ by $(x-1)$ : $r(x) = (x-1)(x^2 - 6x + 8)$ Now factor the quadratic $x^2 - 6x + 8$ . You need two numbers that multiply to 8 and add to -6: those are -2 and -4. So $x^2 - 6x + 8 = (x-2)(x-4)$ . Therefore: $r(x) = (x-1)(x-2)(x-4)$ This gives us three distinct zeros: $x = 1, 2, 4$ , each with multiplicity 1. When a zero has odd multiplicity, the graph crosses the x-axis at that point. Answer A stops at the quadratic without factoring it completely. Answer B incorrectly claims $(x-4)^2$ is a factor, which would mean $x = 4$ is a repeated zero, but our factorization shows it has multiplicity 1. Answer C makes the same error with $(x-1)^2$ , incorrectly suggesting $x = 1$ is a repeated zero. The key strategy here is to always factor completely and remember that odd multiplicities mean the graph crosses the x-axis, while even multiplicities mean it touches and turns around. Don't stop factoring until you've broken down all quadratics into linear factors when possible.
Use zeros to construct a rough graph of $r(x)=x(x-3)(x+2)^2.$ Identify all zeros with multiplicities, indicate whether the graph crosses or touches at each x-intercept, and state the end behavior.
Answer: Zeros: $x=0$ (mult. 1), $x=3$ (mult. 1), $x=-2$ (mult. 2). Crosses at $0$ and $3$ ; touches at $-2$ . End behavior: both ends up. This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For r(x) = x (x - 3) $(x + 2)^2$ , the zeros are x = 0 (multiplicity 1, from x = 0), x = 3 (multiplicity 1, from x - 3 = 0), and x = -2 (multiplicity 2, from $(x + 2)^2$ = 0); the graph crosses at x = 0 and x = 3 due to odd multiplicity and touches at x = -2 due to even multiplicity, with degree 4 (even) and positive leading coefficient indicating both ends up. Choice A correctly identifies the zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice B fails because it states crossing at all zeros, ignoring the even multiplicity at x = -2 that causes touching instead. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!
What is the end behavior of $p(x)=3(x-2)(x+1)(x-4)^2?$ (You may also use the degree and leading coefficient to decide.)
Answer: As $x\to-\infty$ , $p(x)\to\infty$ and as $x\to\infty$ , $p(x)\to\infty$ . This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For $p(x) = 3(x-2)(x+1)(x-4)^2$ , the degree is $1+1+2=4$ (even). When expanded, the leading term will be $3x^4$ , which has positive coefficient. For even degree with positive leading coefficient, both ends go up: as $x \to -\infty$ , $p(x) \to \infty$ and as $x \to \infty$ , $p(x) \to \infty$ . Choice C correctly states this end behavior. Choice A shows the end behavior for odd degree (opposite ends), Choice B shows odd degree with negative leading coefficient, and Choice D shows even degree with negative leading coefficient. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor ( $x - r$ ) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need!
Use zeros to construct a rough graph of $p(x)=(x+2)^2(x-3).$ Identify the zeros (with multiplicities), describe whether the graph crosses or touches the x-axis at each zero, and state the end behavior.
Answer: Zeros: $x=-2$ (mult. 2), $x=3$ (mult. 1). Touches at $x=-2$ , crosses at $x=3$ . End behavior: as $x\to-\infty$ , $p(x)\to-\infty$ and as $x\to\infty$ , $p(x)\to\infty$ . This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For p(x) = (x+2)²(x-3), we have zeros at x = -2 (multiplicity 2, from (x+2)²) and x = 3 (multiplicity 1, from (x-3)). Since multiplicity 2 is even, the graph touches at x = -2; since multiplicity 1 is odd, it crosses at x = 3. The polynomial has degree 3 (odd) with positive leading coefficient, so as x→-∞, p(x)→-∞ and as x→∞, p(x)→∞. Choice A correctly identifies zeros x = -2 (mult. 2) and x = 3 (mult. 1), states the graph touches at x = -2 and crosses at x = 3, and has the correct end behavior. Choice B incorrectly identifies the zero from (x+2)² as x = 2 instead of x = -2. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need!
For $q(x)=-2(x-2)^2(x+4),$ identify the zeros and determine crossing vs. touching at each zero. Also state the end behavior.
Answer: Zeros: $x=2$ (mult. 2), $x=-4$ (mult. 1). Touches at $2$ ; crosses at $-4$ . End behavior: left up, right down. This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For q(x) = -2 $(x - 2)^2$ (x + 4), the zeros are x = 2 (multiplicity 2, from $(x - 2)^2$ = 0) and x = -4 (multiplicity 1, from x + 4 = 0); the graph touches at x = 2 due to even multiplicity and crosses at x = -4 due to odd multiplicity, with degree 3 (odd) and negative leading coefficient (-2) indicating left up and right down. Choice A correctly identifies the zeros with multiplicities, shows proper touching and crossing, and has correct end behavior. Choice D fails because it incorrectly states both ends up, which would be for a positive leading coefficient with even degree, but here it's odd degree and negative. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!
Identify the zeros and their multiplicities for $p(x)=x^2(x-4)(x+1)^2$ and determine whether the graph crosses or touches at each zero.
Answer: Zeros: $x=0$ (mult. 2), $x=4$ (mult. 1), $x=-1$ (mult. 2). Touches at $0$ and $-1$ , crosses at $4$ . This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. For p(x) = x²(x-4)(x+1)², we identify zeros: x²=0 gives x=0 (multiplicity 2), x-4=0 gives x=4 (multiplicity 1), and (x+1)²=0 gives x=-1 (multiplicity 2). At x=0 and x=-1, the multiplicities are even (2), so the graph touches and bounces. At x=4, the multiplicity is odd (1), so the graph crosses. Choice A correctly identifies zeros x=0 (mult. 2), x=4 (mult. 1), x=-1 (mult. 2) and states the graph touches at 0 and -1, crosses at 4. Choice C incorrectly identifies x=-4 instead of x=4 as a zero, misreading the factor (x-4). The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!
Identify zeros and their multiplicities for $p(x)=(x-5)(x+1)^2(x-2)^2,$ then determine the end behavior and whether the graph crosses or touches at each zero.
Answer: Zeros: $x=5$ (mult. 1), $x=-1$ (mult. 2), $x=2$ (mult. 2). Crosses at $5$ , touches at $-1$ and $2$ . End behavior: as $x\to-\infty$ , $p(x)\to-\infty$ and as $x\to\infty$ , $p(x)\to\infty$ . This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For p(x) = (x-5)(x+1)²(x-2)², we have zeros at x = 5 (multiplicity 1, from (x-5)), x = -1 (multiplicity 2, from (x+1)²), and x = 2 (multiplicity 2, from (x-2)²). The graph crosses at x = 5 (odd multiplicity) and touches at both x = -1 and x = 2 (even multiplicities). The polynomial has degree 1+2+2=5 (odd) with positive leading coefficient, so as x→-∞, p(x)→-∞ and as x→∞, p(x)→∞. Choice A correctly identifies all three zeros with their multiplicities, states that the graph crosses at 5 and touches at -1 and 2, and has the correct end behavior for an odd-degree polynomial with positive leading coefficient. Choice B incorrectly claims the graph touches at all three zeros, which would require all multiplicities to be even. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need!
A polynomial has factored form $p(x)=3(x+2)^2(x-1)(x-4).$ Sketch $p(x)$ showing x-intercepts and end behavior. Which option correctly lists the intercept behavior and end behavior?
Answer: x-intercepts at $x=-2,1,4$ . Touches at $x=-2$ and crosses at $x=1$ and $x=4$ . End behavior: both ends up. This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For p(x) = 3 $(x + 2)^2$ (x - 1)(x - 4), zeros are x = -2 (multiplicity 2, touches), x = 1 (multiplicity 1, crosses), x = 4 (multiplicity 1, crosses), degree 4 even positive (leading 3 > 0), both ends up. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice C mistakes the touching zero, but multiplicity 2 at x = -2 means touch, not at x = 1 which is odd—check exponents for each factor! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!
How does the graph behave at each zero of $g(x)=(x-3)^2(x+1)^3?$ Include multiplicities and the end behavior.
Answer: Zeros: $x=3$ (mult. 2) and $x=-1$ (mult. 3). Touches at $x=3$ ; crosses (flattened) at $x=-1$ . End behavior: left down, right up. This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For g(x) = $(x - 3)^2$ $(x + 1)^3$ , zeros are x = 3 (multiplicity 2, touches), x = -1 (multiplicity 3, crosses flattened), degree 5 odd positive, left down right up. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice B swaps the behaviors, but even multiplicity touches and odd crosses—use the 'odd crossers, even bouncers' aid to remember! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!
For $p(x)=x(x-2)^2(x+3),$ identify zeros with multiplicities, determine whether the graph crosses or touches the x-axis at each zero, and state the end behavior.
Answer: Zeros: $x=0$ (mult. 1), $x=2$ (mult. 2), $x=-3$ (mult. 1). Crosses at $0$ and $-3$ , touches at $2$ . End behavior: as $x\to-\infty$ , $p(x)\to\infty$ and as $x\to\infty$ , $p(x)\to\infty$ . This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For p(x) = x(x-2)²(x+3), we have zeros at x = 0 (multiplicity 1, from x), x = 2 (multiplicity 2, from (x-2)²), and x = -3 (multiplicity 1, from (x+3)). The graph crosses at x = 0 and x = -3 (odd multiplicities) and touches at x = 2 (even multiplicity). The polynomial has degree 1+2+1=4 (even) with positive leading coefficient, so both ends go up: as x→-∞, p(x)→∞ and as x→∞, p(x)→∞. Choice A correctly identifies all three zeros with their multiplicities, states that the graph crosses at 0 and -3 while touching at 2, and has the correct end behavior for an even-degree polynomial with positive leading coefficient. Choice B incorrectly claims x = 0 has multiplicity 2, which would make the graph touch there instead of cross. The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need!
Sketch $f(x)=(x+4)(x+1)(x-2)^2$ showing all zeros, their multiplicities, and the correct end behavior. Which description matches?
Answer: Zeros: $x=-4$ (mult. 1), $x=-1$ (mult. 1), $x=2$ (mult. 2). Crosses at $-4$ and $-1$ , touches at $2$ . End behavior: both ends up. This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! For f(x) = (x + 4)(x + 1) $(x - 2)^2$ , zeros are x = -4 (multiplicity 1, crosses), x = -1 (multiplicity 1, crosses), x = 2 (multiplicity 2, touches), degree 4 even positive, both ends up. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice B misidentifies zeros like x = 4 instead of x = -4—always solve factors with attention to signs! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!
Factor the polynomial $p(x)=x^4-5x^2+4$ and use the zeros to sketch a rough graph (x-intercepts, crossing vs touching, and end behavior). Which option is correct?
Answer: Factored form: $(x-2)(x+2)(x-1)(x+1)$ . Zeros: $-2,-1,1,2$ (all mult. 1). Crosses at each intercept. End behavior: both ends up. This question tests your ability to identify zeros from a polynomial's factored form and use them, along with multiplicity information and end behavior, to construct a rough sketch of the polynomial's graph. Zeros from factored form p(x) = a(x - r₁)(x - r₂)... are found by setting each factor equal to zero: from (x - r), the zero is x = r. Multiplicity (how many times a factor appears) determines behavior at that zero: odd multiplicity means the graph crosses the x-axis, even multiplicity means it touches and bounces back. For example, (x - 2)² makes the graph touch at x = 2, while (x - 2)³ makes it cross but with a flattened shape. End behavior depends only on the leading term (highest degree): for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree, ends go opposite directions (positive leading coefficient: left down, right up). This plus the zeros gives you the skeleton of the graph! First factor $x^4$ - $5x^2$ + 4 as ( $x^2$ - 4)( $x^2$ - 1) = (x - 2)(x + 2)(x - 1)(x + 1), so zeros x = -2, -1, 1, 2 (all multiplicity 1, crosses each), degree 4 even positive, both ends up. Choice A correctly identifies zeros with multiplicities, shows proper crossing and touching, and has correct end behavior. Choice B uses an incorrect factoring like ( $x^2$ - 4)( $x^2$ + 1), which gives imaginary zeros and wrong polynomial—double-check factoring by expanding to verify! The four-step polynomial sketching strategy: (1) Find all zeros by setting each factor (x - r) equal to zero (watch signs!), (2) Determine multiplicity of each zero (count how many times factor appears) and whether graph crosses (odd) or touches (even), (3) Find end behavior using degree (even = same both ends, odd = opposite ends) and leading coefficient sign (positive eventually goes up, negative eventually goes down), (4) Mark zeros on x-axis and connect with smooth curve showing proper behavior at each zero and correct end behavior. Rough shape is all you need! Multiplicity memory aid: Think 'odd crossers, even bouncers.' Odd multiplicity (1, 3, 5...) = graph crosses through the x-axis. Even multiplicity (2, 4, 6...) = graph bounces off the x-axis without crossing. Higher multiplicity = flatter at that zero. So (x - 3)² touches and turns around, (x - 3)³ crosses but flattens, (x - 3)⁴ touches with even more flattening. The pattern is consistent!

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