Find a Polynomial Given its Graph: Questions with Solutions

Step 1: Identify Zeros and Multiplicities
The graph of the function crosses the x-axis cleanly at \( x = -1 \). This indicates a zero of multiplicity 1, corresponding to the factor \( (x + 1) \).
The graph touches the x-axis and turns around at \( x = 3 \). This indicates a zero of multiplicity 2 (a parabolic shape), corresponding to the factor \( (x - 3)^2 \).

Step 2: Set up the Equation
Based on the factors, the function \( g \) has the general form:
\[ g(x) = k(x + 1)(x - 3)^2 \] where \( k \) is an unknown constant multiplier.

Step 3: Solve for \( k \)
We can find \( k \) using the marked point on the graph with coordinates \( (1, 3) \):
\[ g(1) = k(1 + 1)(1 - 3)^2 = 3 \] \[ k(2)(-2)^2 = 3 \] \[ 8k = 3 \implies k = \dfrac{3}{8} \]

Conclusion:
The exact equation for \( g(x) \) is:
\[ g(x) = \dfrac{3}{8}(x + 1)(x - 3)^2 \]

Step 1: Identify Zeros and Multiplicities
By analyzing the x-intercepts, we find three distinct points of interaction:
* A straight cross at \( x = 2 \), meaning a multiplicity of 1, giving factor \( (x - 2) \).
* A straight cross at \( x = -2 \), meaning a multiplicity of 1, giving factor \( (x + 2) \).
* A touch-and-turn at \( x = -1 \), meaning a multiplicity of 2, giving factor \( (x + 1)^2 \).

Step 2: Set up the Equation
Multiplying these factors yields our base 4th degree polynomial:
\[ f(x) = k(x - 2)(x + 2)(x + 1)^2 \] where \( k \) is a constant.

Step 3: Solve for \( k \)
The graph reveals a clear y-intercept at \( (0, -1) \). Substitute \( x = 0 \) and \( y = -1 \):
\[ f(0) = k(0 - 2)(0 + 2)(0 + 1)^2 = -1 \] \[ k(-2)(2)(1)^2 = -1 \] \[ -4k = -1 \implies k = \dfrac{1}{4} \]

Conclusion:
The equation is:
\[ f(x) = \dfrac{1}{4}(x - 2)(x + 2)(x + 1)^2 \]

Step 1: Identify Zeros and Multiplicities
The graph has x-intercepts at \( x = 0 \) and \( x = \dfrac{5}{2} \). Because the graph crosses the x-axis at both points, both zeros must have an odd multiplicity.
* At \( x = 0 \), the graph exhibits a flattened "S" or cubic shape. This indicates a multiplicity of 3, yielding the factor \( x^3 \).
* At \( x = \dfrac{5}{2} \), the cross is direct and relatively linear, indicating a multiplicity of 1, yielding the factor \( \left(x - \dfrac{5}{2}\right) \).

Step 2: Set up the Equation
Using these observations, we write the general form:
\[ f(x) = k(x)^3 \left(x - \dfrac{5}{2}\right) \]

Step 3: Solve for \( k \)
We use the given reference point \( (2, -4) \) to find \( k \):
\[ -4 = k(2)^3 \left(2 - \dfrac{5}{2}\right) \] \[ -4 = k(8) \left(-\dfrac{1}{2}\right) \] \[ -4 = -4k \implies k = 1 \]

Conclusion:
The equation of the polynomial is:
\[ f(x) = x^3 \left(x - \dfrac{5}{2}\right) \]

Step 1: Factored Form Setup
The polynomial is degree 3. The graph shows a zero of multiplicity 1 at \( x = -2 \) (factor: \( x + 2 \)), and a zero of multiplicity 2 at \( x = 1 \) (factor: \( (x - 1)^2 \)).
\[ y = k(x + 2)(x - 1)^2 \]

Step 2: Find \( k \)
Use the clear y-intercept at \( (0, 1) \):
\[ 1 = k(0 + 2)(0 - 1)^2 \] \[ 1 = 2k \implies k = \dfrac{1}{2} = 0.5 \]

Step 3: Expand to Standard Form
Now, expand the factored polynomial to reveal the individual coefficients:
\[ y = 0.5(x + 2)(x^2 - 2x + 1) \] \[ y = 0.5(x^3 - 2x^2 + x + 2x^2 - 4x + 2) \] \[ y = 0.5(x^3 - 3x + 2) \] \[ y = 0.5x^3 - 1.5x + 1 \]

Conclusion:
Comparing this to \( y = ax^3 + bx^2 + cx + d \), we deduce:
\[ a = 0.5, \quad b = 0, \quad c = -1.5, \quad d = 1 \]

Step 1: Leverage Symmetry Concepts
Since the graph is perfectly symmetric across the y-axis, the function must be an even function (where \( f(x) = f(-x) \)).
For a polynomial to be an even function, it can only contain even powers of \( x \). The terms \( bx^3 \) and \( dx \) are odd functions. For the overall polynomial to remain even, their coefficients must be zero:
\[ b = 0, \quad d = 0 \]

Step 2: Find the Constant \( e \)
The polynomial simplifies to \( y = a x^4 + c x^2 + e \).
Coefficient \( e \) represents the y-intercept. Examining the graph, the y-intercept is clearly at \( (0, -2) \):
\[ -2 = a(0)^4 + c(0)^2 + e \] \[ e = -2 \]

Conclusion:
The required coefficients are:
\[ b = 0, \quad d = 0, \quad e = -2 \]

Step 1: Factor Construction
* Crosses directly at \( x = -2 \implies \) Multiplicity 1: factor \( (x + 2) \).
* Tangent at \( x = 1 \implies \) Multiplicity 2: factor \( (x - 1)^2 \).
* Tangent at \( x = 4 \implies \) Multiplicity 2: factor \( (x - 4)^2 \).
(Check the degree: 1 + 2 + 2 = 5. This aligns perfectly with the stated degree.)

Step 2: Setting up the Equation
\[ P(x) = k(x + 2)(x - 1)^2(x - 4)^2 \]

Step 3: Solving for \( k \)
Use the y-intercept \( (0, -16) \):
\[ -16 = k(0 + 2)(0 - 1)^2(0 - 4)^2 \] \[ -16 = k(2)(-1)^2(-4)^2 \] \[ -16 = k(2)(1)(16) \] \[ -16 = 32k \implies k = -\dfrac{1}{2} \]

Conclusion:
\[ P(x) = -0.5(x + 2)(x - 1)^2(x - 4)^2 \]

Step 1: Understand Odd Symmetry
An odd function is symmetric about the origin (\( Q(-x) = -Q(x) \)). This means whatever graphical behavior happens on the positive x-axis must be mirrored (diagonally) on the negative x-axis.
* Tangent at \( x = 2 \implies \) Tangent at \( x = -2 \).
* Passes through the origin \( \implies \) Root at \( x = 0 \).

Step 2: Factor Construction (Least Degree)
* Root at \( 0 \) crossing through \( \implies \) Multiplicity 1: factor \( (x) \).
* Tangent at \( 2 \implies \) Multiplicity 2: factor \( (x - 2)^2 \).
* Tangent at \( -2 \implies \) Multiplicity 2: factor \( (x + 2)^2 \).
(The least degree is 1 + 2 + 2 = 5).

Step 3: Solving for \( k \)
\[ Q(x) = k(x)(x - 2)^2(x + 2)^2 \]
Use the point \( (1, -9) \):
\[ -9 = k(1)(1 - 2)^2(1 + 2)^2 \] \[ -9 = k(1)(-1)^2(3)^2 \] \[ -9 = k(1)(1)(9) \] \[ 9k = -9 \implies k = -1 \]

Conclusion:
\[ Q(x) = -x(x - 2)^2(x + 2)^2 \]

Step 1: Translating Shape to Multiplicity
* Parabolic touch at \( x = -3 \implies \) Even multiplicity. Least degree = 2. Factor: \( (x + 3)^2 \).
* Straight line cross at \( x = 2 \implies \) Multiplicity 1. Factor: \( (x - 2) \).
* Flat "S" shape cross at \( x = 5 \implies \) Odd multiplicity greater than 1. Least degree = 3. Factor: \( (x - 5)^3 \).

Step 2: Setting up the Equation
Multiplying these factors creates a 6th degree polynomial:
\[ H(x) = k(x + 3)^2(x - 2)(x - 5)^3 \]

Step 3: Solving for \( k \)
Substitute the point \( (3, -72) \):
\[ -72 = k(3 + 3)^2(3 - 2)(3 - 5)^3 \] \[ -72 = k(6)^2(1)(-2)^3 \] \[ -72 = k(36)(-8) \] \[ -72 = -288k \]
\[ k = \dfrac{-72}{-288} = \dfrac{1}{4} \]

Conclusion:
\[ H(x) = 0.25(x + 3)^2(x - 2)(x - 5)^3 \]