Math Problem Statement

The quartic polynomial f(x) = x^4 + x^3 - 28x^2 + 20x + 48 is shown graphed. (a) Why do the ends of this polynomial point in the same direction compared to the cubic polynomials that pointed in opposite directions? (b) Suggest a transformation of this function that would cause the ends to point downward but would leave the zeros unchanged. (c) Circle the four real zeros this function has. List them below. (d) Explain how vertically shifting this function could produce a quartic with no real zeros at all.

Solution

Let’s analyze the given problem and answer each part step by step:


Given Polynomial:

f(x)=x4+x328x2+20x+48f(x) = x^4 + x^3 - 28x^2 + 20x + 48


Part (a): Why do the ends of this polynomial point in the same direction?

The polynomial f(x)f(x) is a quartic (degree 4) polynomial. For a polynomial, the behavior of the ends is determined by the leading term. Here, the leading term is x4x^4, which has a positive coefficient. Quartic polynomials with positive leading terms have their ends pointing in the same direction (upward). This is unlike cubic polynomials, whose odd-degree nature causes their ends to point in opposite directions.


Part (b): Transformation to make the ends point downward without changing the zeros

To make the ends point downward, we need to negate the leading term of f(x)f(x). This can be achieved by multiplying the entire polynomial by 1-1, giving: f(x)f(x)=(x4+x328x2+20x+48).f(x) \to -f(x) = -(x^4 + x^3 - 28x^2 + 20x + 48). This transformation flips the graph vertically, reversing the end behavior, but it does not affect the zeros of the polynomial.


Part (c): Circle the four real zeros and list them

From the graph, the four real zeros appear to be approximately at: x=6,x=2,x=2,x=4.x = -6, \, x = -2, \, x = 2, \, x = 4.


Part (d): How vertically shifting the function could remove real zeros

By vertically shifting the graph of f(x)f(x), we add or subtract a constant to/from the polynomial: f(x)+k,f(x) + k, where kk is a constant.

  • Effect of vertical shifting on zeros: Adding a constant kk can move the entire graph up or down. If kk is large enough (positive or negative), the graph may no longer intersect the x-axis. When this happens, the polynomial would have no real zeros.

For instance, shifting f(x)f(x) downward (subtracting a large kk) could cause the local maxima and minima to drop below the x-axis, eliminating all real roots.


Would you like to explore the answers in more detail or see specific calculations? Here are related questions for deeper understanding:

  1. How does the degree of a polynomial influence the end behavior of its graph?
  2. What transformations can preserve the shape of the graph but affect the direction of its ends?
  3. How can we algebraically confirm the approximate zeros using synthetic division or the Rational Root Theorem?
  4. What happens to the number of real zeros if we vertically stretch or compress the graph?
  5. Can you describe the role of the discriminant in identifying the nature of the roots?

Tip: When analyzing transformations of functions, always focus on how they affect key features like intercepts, zeros, and end behavior!

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Math Problem Analysis

Mathematical Concepts

Polynomial Graphs
End Behavior
Function Transformations
Zeros of Polynomials

Formulas

f(x) = x^4 + x^3 - 28x^2 + 20x + 48

Theorems

Fundamental Theorem of Algebra

Suitable Grade Level

Grades 10-12