Problem 72 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine whether the statement is true or false. Justify your answer. The function \(f(x)=a(x-5)^{2}\) has exactly one \(x\) -intercept for any nonzero value of \(a\)

Step-by-Step Solution

Answer

The statement is true. The function has exactly one x-intercept (at \(x=5\)) for any nonzero value of \(a\).

Step 1: Understand the problem

The function given is a quadratic equation, which graphically represents a parabola. The x-intercept(s) of a function are the point(s) where the graph of the function crosses or touches the x-axis.

Step 2: Determine the x-intercept

For finding x-intercepts, we can set \(f(x)\) equal to zero and solve for \(x\), because the y-coordinate at any x-intercept is 0.\nSo, set \(f(x)=0\) => \(a(x-5)^{2}=0\).\nThis equation is equal to zero only when \(x=5\), regardless of the value of \(a\) (as long as \(a\) is nonzero).

Step 3: Conclusion

So, the function \$f(x)=a(x-5)^{2}\$ has exactly one x-intercept (at \(x=5\)) for any nonzero value of \(a\). Therefore, the statement is true.

Key Concepts

Parabola Graph

A parabola is a symmetrical, U-shaped graph that is the visual representation of a quadratic equation. It's essential to visualize this because it helps in understanding the behavior of quadratic functions. The general form of a quadratic function is \( f(x) = ax^2 + bx + c \), where \( a, b, \text{and} c \) are constants and \( a \) is nonzero. For the function \( f(x)=a(x-5)^{2} \), the graph is a parabola that opens upwards if \( a > 0 \) and downwards if \( a < 0 \) due to the square term.

A unique feature of a parabola is its vertex, the highest or lowest point depending on the direction it opens. For \( f(x)=a(x-5)^{2} \), the vertex is at (5, 0). The axis of symmetry is a vertical line that goes through the vertex, and for this function, it is \( x=5 \). Since a parabola is symmetric, the x-intercepts (if they exist) occur at equal distances from the axis of symmetry.

X-Intercept Calculation

To find the x-intercept(s) of a quadratic function, we set \( f(x) \) equal to zero and solve for \( x \) since the y-coordinate of an x-intercept is always 0. For the given function \( f(x)=a(x-5)^{2} \) we set \( a(x-5)^{2}=0 \).Here's the calculation:\[ 0 = a(x - 5)^2 \]Since \( a \) is nonzero, we can divide both sides by \( a \) without changing the equation:\[ 0 = (x - 5)^2 \]To solve for \( x \), we take the square root of both sides:\[ \sqrt{0} = \sqrt{(x - 5)^2} \]Hence, the solution for \( x \) is 5, meaning the graph intersects the x-axis at the point \( (5, 0) \) only, making it the sole x-intercept.

Quadratic Function Properties

Quadratic functions exhibit several inherent properties that govern their shape and position on a graph.

• Direction: The sign of \( a \) determines whether the parabola opens upwards (\( a > 0 \) or downwards (\( a < 0 \).
• Vertex: This is the point where the parabola either reaches a maximum (if it opens downwards) or minimum (if it opens upwards). The vertex for \( f(x)=a(x-5)^{2} \) is at the point \( (5, 0) \).
• Axis of symmetry: For the function at hand, the axis of symmetry is the vertical line \( x=5 \).
• Y-intercept: This is the point where the graph crosses the y-axis, which can be found by substituting \( x=0 \) into the function.
• X-intercepts: The points where the graph crosses the x-axis. A quadratic function can have 0, 1, or 2 x-intercepts.
• Domain and Range: The domain of a quadratic function is all real numbers, whereas the range is all values \( y \) such that \( y \geq c \) (if \( a > 0 \) or \( y \leq c \) (if \( a < 0 \) where \( c \) is the y-coordinate of the vertex.

Each of these properties plays a role in defining the precise shape and position of the parabola on a coordinate plane.

Solving Quadratic Equations

The process of solving quadratic equations involves finding the values of \( x \) for which \( ax^2 + bx + c = 0 \) holds true. There are several methods to do this:

• Factoring: Expressing the quadratic as a product of binomials, if possible, and using the zero product property.
• Completing the square: Rewriting the quadratic in the form \( (x-h)^2 = k \) to find the vertex form.
• The quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) which derives from the process of completing the square and provides a systematic solution.
• Graphing: Finding the x-intercepts of the function on the coordinate plane if they exist.

In the context of the exercise, since \( f(x) \) is already in vertex form \( a(x-h)^{2}+k\), we find the x-intercept by setting \( f(x) \) to zero and solving for \( x \) which simplifies to finding the square root of both sides after isolating the \( (x-h)^2 \) term. It's crucial to note that not all quadratic equations will factor neatly or have real solutions, but the quadratic formula always provides an answer, either real or complex.