Graphs of Polynomials - Graphing & Polynomial creation

Question 1: Matching an Equation to its Graph (from file image_d096bd.png)

The Question: Which of the following represents the graph of the polynomial $p(x) = \frac{1}{2}(x-1)^2(x+1)^3(x-2)$?

Detailed Solution:

Let’s break down the equation to predict the graph’s behavior.

1. Find the Zeros: The zeros are the x-values that make the factors zero.

   • From $(x-1)^2 \implies x = 1$
   • From $(x+1)^3 \implies x = -1$
   • From $(x-2) \implies x = 2$ The zeros are at -1, 1, and 2.

2. Determine the Multiplicity at each Zero:

   • At x = 1: The factor is $(x-1)^2$. The multiplicity is 2 (even), so the graph must touch the x-axis and bounce.
   • At x = -1: The factor is $(x+1)^3$. The multiplicity is 3 (odd), so the graph must cross the x-axis (and flatten as it does).
   • At x = 2: The factor is $(x-2)^1$. The multiplicity is 1 (odd), so the graph must cross the x-axis.

3. Determine the End Behavior:

   • The degree of the polynomial is the sum of the multiplicities: $2 + 3 + 1 = 6$. This is an even degree.
   • The leading coefficient is $\frac{1}{2}$, which is positive.
   • An even-degree polynomial with a positive leading coefficient has “Up-Up” behavior (it rises on the far left and rises on the far right).

4. Match with the Graphs:

   • We need a graph that crosses at x=-1 and x=2, bounces at x=1, and goes up on both ends.
   • Only the second graph from the top satisfies all these conditions.

Final Answer: The second graph is the correct representation.

Question 2: Intermediate Value Theorem (from file image_d093f2.png)

The Question: If $p(x)$ is a polynomial function and $p(3)$ and $p(5)$ have opposite signs to each other, then which of the following is true?

Core Concept: This is a direct application of the Intermediate Value Theorem. A polynomial graph is a continuous, unbroken curve.

Detailed Solution:

1. We are given that one of the values, $p(3)$ or $p(5)$, is positive, and the other is negative.
2. Imagine these two points on a graph. One point is above the x-axis, and the other is below it.
3. Since a polynomial’s graph is continuous, to get from the point above the axis to the point below the axis, the curve must cross the x-axis at some point in between.
4. A point where the graph crosses the x-axis is a root, or a value $a$ where $p(a)=0$.
5. This crossing point $a$ must be located between $x=3$ and $x=5$.

Final Answer: There exists, at least one value $a$ such that $3 < a < 5$ and $p(a) = 0$.

Question 3: Finding an Equation from a Graph (from file image_d093f2.png)

The Question: The algebraic expression for the polynomial of the graph as shown in the Figure AQ-7.8 is __________.

Core Concept: We can reconstruct the factored form of a polynomial by identifying its zeros, their multiplicities, and using another point (like the y-intercept) to find the stretch factor, $a$.

Detailed Solution:

The provided graph has some inconsistencies. Let’s analyze what the graph shows versus what the options imply.

Analysis of the Graph’s Features:

1. Zeros: The graph intersects the x-axis at $x=-1$, $x=1$, and $x=2$.
2. Multiplicities:
   • At $x=-1$, the graph crosses (odd multiplicity, likely 1). Factor: $(x+1)$.
   • At $x=1$, the graph touches and bounces (even multiplicity, likely 2). Factor: $(x-1)^2$.
   • At $x=2$, the graph crosses (odd multiplicity, likely 1). Factor: $(x-2)$.
3. End Behavior: The graph goes up on the far left and up on the far right (“Up-Up”). This implies an even degree and a positive leading coefficient ($a>0$). The sum of our likely multiplicities is $1+2+1=4$, which is an even degree, consistent with the end behavior.
4. Find Stretch Factor ‘a’: The graph passes through the y-intercept $(0, 1)$. Let’s use the factored form we derived: $p(x) = a(x+1)(x-1)^2(x-2)$.
   • $1 = a(0+1)(0-1)^2(0-2)$
   • $1 = a(1)(1)(-2) \implies 1 = -2a \implies a = -1/2$.

The Contradiction: Our analysis of the end behavior shows that the stretch factor ‘$a$’ must be positive. However, using the y-intercept to calculate it gives $a=-1/2$. This means the graph is drawn with mathematically inconsistent features.

Given the options, it seems the intended question might have ignored the y-intercept. However, none of the options perfectly match the multiplicities shown in the graph. The question is flawed.

Final Answer: Based on the visual information, the graph is impossible as drawn and does not match any of the provided options correctly.

Question 4: Even and Odd Polynomials (from file image_d093d6.png)

The Question: Given three polynomial are $p_1(x) = (-x^2 + x^4)^2$, $p_2(x) = -x^3 + 5x$, $p_3(x) = 9x^3 + 7x$, choose the set of correct options.

Core Concept: A polynomial is even if all its terms have even exponents. It’s odd if all its terms have odd exponents. It’s neither if it has a mix.

Detailed Solution:

1. Analyze $p_1(x) = (-x^2 + x^4)^2$:

   • Let’s expand this: $(-x^2 + x^4)(-x^2 + x^4) = (-x^2)(-x^2) + 2(-x^2)(x^4) + (x^4)(x^4)$
   • $= x^4 - 2x^6 + x^8$.
   • The powers are 4, 6, and 8. Since all exponents are even, $p_1(x)$ is an even polynomial.

2. Analyze $p_2(x) = -x^3 + 5x$:

   • The terms are $-x^3$ and $5x^1$.
   • The powers are 3 and 1. Since all exponents are odd, $p_2(x)$ is an odd polynomial.

3. Analyze $p_3(x) = 9x^3 + 7x$:

   • The terms are $9x^3$ and $7x^1$.
   • The powers are 3 and 1. Since all exponents are odd, $p_3(x)$ is an odd polynomial.

Now evaluate the statements:

• “$p_1, p_2$ are even polynomials”: FALSE ($p_2$ is odd).
• “$p_3$ is an odd polynomial”: TRUE.
• “$p_1$ is an even polynomial”: TRUE.
• “$p_3$ is an even polynomial”: FALSE.
• “$p_1, p_2$ are odd polynomials”: FALSE ($p_1$ is even).

Final Answer: The correct options are:

• $p_3$ is an odd polynomial
• $p_1$ is an even polynomial

Question 5: Stretch Factor of a Graph (from file image_d093b6.png)

The Question: Stretch factor of the given graph as in the Figure AQ-7.9 of the polynomial is __________.

Core Concept: The “stretch factor” is the leading coefficient ‘$a$’ in the factored form of the polynomial, $p(x) = a(x-r_1)(x-r_2)…$.

Detailed Solution:

1. Identify the Zeros (Roots) from the graph:
   • The graph crosses the x-axis at three points: $x=-1$, $x=1$, and $x=2$.
2. Determine the Multiplicity:
   • At each of these zeros, the graph crosses the axis directly without bouncing or flattening. This suggests each zero has a multiplicity of 1.
3. Write the partial factored form:
   • $p(x) = a(x - (-1))(x - 1)(x - 2)$
   • $p(x) = a(x+1)(x-1)(x-2)$
4. Find a convenient point on the graph to solve for ‘a’:
   • The graph clearly passes through the y-intercept at the point $(0, 8)$.
5. Substitute the coordinates of this point into the equation:
   • Let $x=0$ and $p(x)=8$.
   • $8 = a(0+1)(0-1)(0-2)$
   • $8 = a(1)(-1)(-2)$
   • $8 = a(2)$
6. Solve for a:
   • $a = \frac{8}{2} = 4$

Final Answer: The stretch factor is 4.