This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. When a polynomial is in factored form like P(x) = a(x - r₁)(x - r₂)(x - r₃), the zeros are immediately visible: set each factor equal to zero to get x = r₁, r₂, r₃. These zeros are the x-intercepts—points where the graph crosses or touches the x-axis—and they're the foundation for sketching the graph because they divide the x-axis into regions where the polynomial is positive or negative. From the factored form P(x) = (x-1)(x-5)(x+2), we find zeros by setting each factor equal to zero: (x-1) = 0 → x = 1, (x-5) = 0 → x = 5, (x+2) = 0 → x = -2. The zeros (break-even points) are x = -2, 1, 5. Remember: from (x - r), the zero is x = r (opposite sign!), so (x - 3) gives zero at x = 3, and (x + 5) = (x - (-5)) gives zero at x = -5. Choice A correctly identifies break-even x-values as -2, 1, 5 and shows x-intercepts as (-2,0), (1,0), (5,0) by properly applying the zero product property to find where profit equals zero. Choice B has the signs wrong on all zeros: from the factor (x-1), the zero is x = 1, not x = -1. This sign flip is super common! Remember: (x - r) gives zero at x = r, so you reverse the sign from what's in the factor. Think: what value makes (x [sign] [number]) equal to zero? The zero-finding procedure from factored form: for each factor (x - r), set it equal to zero and solve: (x - r) = 0 → x = r. That r is your zero. Do this for every factor. Watch signs carefully: (x - 3) gives x = 3, (x + 5) gives x = -5. If a factor appears multiple times like (x - 2)³, that zero has multiplicity 3. List all zeros (with multiplicities if relevant), and you're ready to sketch!

This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. When a polynomial is in factored form like P(x) = a(x - r₁)(x - r₂), the zeros are immediately visible: set each factor equal to zero to get x = r₁, r₂. These zeros are the x-intercepts—points where the graph crosses or touches the x-axis—and they're the foundation for sketching the graph because they divide the x-axis into regions where the polynomial is positive or negative. From the factored form P(x) = (x - 2)(x - 3), we find zeros by setting each factor equal to zero: (x - 2) = 0 → x = 2, (x - 3) = 0 → x = 3. The zeros are x = 2, 3. Remember: from (x - r), the zero is x = r (opposite sign!), so (x - 2) gives zero at x = 2, but if it were (x + 2), it would give x = -2. Choice A correctly identifies zeros as x=2, 3 and shows the parabola opens up by using positive leading coefficient for both ends up. Choice B has the signs wrong on the zeros: from the factors (x - 2) and (x - 3), the zeros are x = 2 and 3, not x = -2 and -3. This sign flip is super common! Remember: (x - r) gives zero at x = r, so you reverse the sign from what's in the factor. Think: what value makes (x [sign] [number]) equal to zero? The zero-finding procedure from factored form: for each factor (x - r), set it equal to zero and solve: (x - r) = 0 → x = r. That r is your zero. Do this for every factor. Watch signs carefully: (x - 3) gives x = 3, (x + 5) gives x = -5. If a factor appears multiple times like (x - 2)³, that zero has multiplicity 3. List all zeros (with multiplicities if relevant), and you're ready to sketch!

This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. A polynomial's end behavior (what happens as x → ∞ and x → -∞) is determined by its degree and leading coefficient: for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree polynomials, the ends go opposite directions (if positive leading coefficient: left end down, right end up). This end behavior, combined with zeros, gives you the rough shape! This polynomial has degree 4 and leading coefficient positive. Since 4 is even and positive, the end behavior is: as x → -∞ (far left), P(x) → ∞, and as x → ∞ (far right), P(x) → ∞. Think of it this way: even degree means both ends match. This end behavior plus the zeros gives us the skeleton of the graph! Choice A correctly identifies zeros as x=-2, -1, 1, 2 and end behavior both ends up by using degree and leading coefficient. Choice B has the end behavior backwards: with degree 4 (even) and leading coefficient positive, the ends should both up, not left down & right up. Remember even degree means both ends same direction. The sign of the leading coefficient then determines up or down! End behavior memory tricks: Even degree polynomials make 'U-shapes' or 'n-shapes' (both ends same direction), while odd degree polynomials make 'chair shapes' or ' S-curves' (ends opposite). Positive leading coefficient: right end goes up. Negative: right end goes down. Combine these: degree 3 with positive leading coefficient = left down, right up (like sitting in a chair). Degree 4 with negative leading coefficient = both ends down (like an upside-down U). Visual mnemonics help!

This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. A polynomial's end behavior (what happens as x → ∞ and x → -∞) is determined by its degree and leading coefficient: for even-degree polynomials, both ends go the same direction (both up if positive leading coefficient, both down if negative). For odd-degree polynomials, the ends go opposite directions (if positive leading coefficient: left end down, right end up). This end behavior, combined with zeros, gives you the rough shape! This polynomial has degree 3 and leading coefficient negative. Since 3 is odd and negative, the end behavior is: as x → -∞ (far left), P(x) → ∞, and as x → ∞ (far right), P(x) → -∞. Think of it this way: odd degree with negative a means left high, right low. This end behavior plus the zeros gives us the skeleton of the graph! Choice B correctly identifies zeros as x=0, 3, -2 and end behavior left up, right down by using degree and leading coefficient. Choice A has the end behavior backwards: with degree 3 (odd) and leading coefficient negative, the ends should left up & right down, not left down & right up. Remember odd degree means opposite directions. The sign of the leading coefficient then determines up or down! End behavior memory tricks: Even degree polynomials make 'U-shapes' or 'n-shapes' (both ends same direction), while odd degree polynomials make 'chair shapes' or ' S-curves' (ends opposite). Positive leading coefficient: right end goes up. Negative: right end goes down. Combine these: degree 3 with positive leading coefficient = left down, right up (like sitting in a chair). Degree 4 with negative leading coefficient = both ends down (like an upside-down U). Visual mnemonics help!

This question tests your ability to use the zeros (x-intercepts) of a polynomial to sketch its graph and understand how the factored form reveals both where the graph crosses the x-axis and the overall shape of the curve. When a polynomial is in factored form like P(x) = a(x - r₁)(x - r₂)(x - r₃), the zeros are immediately visible: set each factor equal to zero to get x = r₁, r₂, r₃. These zeros are the x-intercepts—points where the graph crosses or touches the x-axis—and they're the foundation for sketching the graph because they divide the x-axis into regions where the polynomial is positive or negative. From the factored form P(x) = (x+2)(x+3), we find zeros by setting each factor equal to zero: (x+2)=0 → x=-2, (x+3)=0 → x=-3. The zeros are x=-2, -3. Remember: from (x - r), the zero is x = r (opposite sign!), so (x - 3) gives zero at x = 3, and (x + 5) = (x - (-5)) gives zero at x = -5. Choice B correctly identifies zeros as x=-2, -3 (x-intercepts (-2,0), (-3,0)) by properly applying zero product property. Choice A has the sign wrong on zeros: from the factors (x+2) and (x+3), the zeros are x=-2 and x=-3, not positive. This sign flip is super common! Remember: (x - r) gives zero at x = r, so you reverse the sign from what's in the factor. Think: what value makes (x [sign] [number]) equal to zero? The zero-finding procedure from factored form: for each factor (x - r), set it equal to zero and solve: (x - r) = 0 → x = r. That r is your zero. Do this for every factor. Watch signs carefully: (x - 3) gives x = 3, (x + 5) gives x = -5. If a factor appears multiple times like (x - 2)³, that zero has multiplicity 3. List all zeros (with multiplicities if relevant), and you're ready to sketch!