Problem 61 - College Algebra

Question

Use the information about the graph of a polynomial function to determine the function. Assume the leading coefficient is 1 or ?1. There may be more than one correct answer. The \(y\) -intercept is \((0,-4) .\) The \(x\) -intercepts are \((-2,0),\) \((2,0) .\) Degree is \(2 .\) End behavior: as \(x \rightarrow-\infty,\) \(f(x) \rightarrow \infty,\) as \(x \rightarrow \infty, f(x) \rightarrow \infty.\)

Step-by-Step Solution

Answer

The polynomial function is \(f(x) = x^2 - 4\).

Step 1: Understand the Intercepts

The given intercepts specify where the graph crosses the axes. The \(y\)-intercept \(0, -4\)\ indicates that when \(x = 0\), \(f(x) = -4\). The \(x\)-intercepts \(-2, 0\) and \(2, 0\)\ show that the polynomial has factors \((x + 2)\) and \((x - 2)\).

Step 2: Consider the Polynomial Degree

The polynomial has a degree of 2, which means it is a quadratic function of the form \(f(x) = a(x + 2)(x - 2)\). This indicates that \(a\) is the leading coefficient to be determined, considering the behavior of the graph.

Step 3: Determine the Leading Coefficient

The end behavior described is as \(x \rightarrow \pm\infty, f(x) \rightarrow \infty\), which corresponds to a parabola opening upwards. Therefore, the leading coefficient \(a\) must be positive, so \(a = 1\).

Step 4: Verify with the Y-Intercept

Substitute \(x = 0\) into \(f(x) = (x + 2)(x - 2)\) and set it equal to the \(y\)-intercept: \[ f(0) = (0 + 2)(0 - 2) = -4 \]This verifies that the polynomial satisfies \(f(0) = -4\).

Step 5: Finalize the Answer

Since all conditions match, the polynomial function is:\[ f(x) = (x + 2)(x - 2) = x^2 - 4 \]

Key Concepts

Y-Intercept

The y-intercept of a function is where the graph of the function crosses the y-axis. It happens when the value of x is zero. This point gives us essential information about how the polynomial behaves when x equals zero.
In the polynomial function provided, the y-intercept is given as \((0, -4)\). This means that when \(x = 0\), \(f(x) = -4\). To verify this when working with the function \(f(x) = (x + 2)(x - 2)\), substitute 0 into the function:

• \(f(0) = (0 + 2)(0 - 2) = -4\)

This confirms the y-intercept, showing where the function crosses the y-axis, at the point \( (0, -4) \). Such intercepts are helpful, especially when sketching graphs, providing an anchor point for the curve.

X-Intercepts

X-intercepts indicate where the graph crosses the x-axis. These are the values of x for which the function equals zero. Finding x-intercepts involves identifying the x-values that make the function's output zero.
For the quadratic function given, the x-intercepts are \((-2, 0)\) and \((2, 0)\). This means \(f(x) = 0\) when \(x = -2\) and \(x = 2\). These intercepts tell us that the factors of the polynomial are \((x + 2)\) and \((x - 2)\).
Because the given polynomial is quadratic, its degree of 2 directly ties into having two x-intercepts. The polynomial is expressed as:

• \(f(x) = (x + 2)(x - 2)\)

Understanding these intercepts is crucial, as they not only help define the specific points where the graph touches or crosses the axis but also lead in forming the factors of the polynomial function.

End Behavior

End behavior describes how the function behaves as the input values become very large or small (as x approaches infinity or negative infinity). For polynomial functions, this behavior is mostly dictated by the degree and the leading coefficient of the polynomial.
In this particular exercise, the end behavior is described as follows:

• As \(x \to -\infty\), \(f(x) \to \infty\).
• As \(x \to \infty\), \(f(x) \to \infty\).

This signifies that both ends of the graph rise, which is characteristic of an upward-opening parabola. The described behavior implies that our quadratic function must have a positive leading coefficient. This also helps in concluding that the leading coefficient \(a = 1\).
When sketching the graph or visualizing it, understanding the end behavior helps anticipate how the graph extends beyond plotted points, ensuring continuity in logical deduction of the polynomial's structure.

Quadratic Function

A quadratic function is a type of polynomial with a degree of 2. This function is typically of the form \(f(x) = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Quadratic functions have distinct characteristics such as a parabolic shape and symmetry.
For our exercise, the quadratic polynomial is determined from the given x-intercepts \(-2\) and \(2\), resulting in a factored form of \(f(x) = (x + 2)(x - 2)\). This can be expanded to:

• \(f(x) = x^2 - 4\)

This function’s graph is a parabola opening upwards, confirmed by the positive leading coefficient.
Quadratics are valuable in modeling many real-world scenarios due to their symmetrical nature and predictable behavior. The roots, which are the x-intercepts, also provide solutions for instances when the function value equals zero, making quadratic functions fundamental in both theoretical and applied mathematics.