Problem 5 - Algebra and Trigonometry

Question

\(5-18=\) A quadratic function is given. (a) Express the quadratic function in standard form. (b) Find its vertex and its \(x-\) and \(y\) -intercept(s). (c) Sketch its graph. $$ f(x)=x^{2}-6 x $$

Step-by-Step Solution

Answer

(a) \( f(x) = x^2 - 6x \). (b) Vertex: (3, -9); X-int: (0, 0) and (6, 0); Y-int: (0, 0). (c) Sketch a parabola opening upwards.

Step 1: Express Quadratic in Standard Form

The given function is already in standard form, which is \( f(x) = ax^2 + bx + c \). For this function, \( a = 1 \), \( b = -6 \), and \( c = 0 \). Therefore, the quadratic function in standard form is \( f(x) = x^2 - 6x \).

Step 2: Find the Vertex

The vertex of a quadratic function in standard form \( f(x) = ax^2 + bx + c \) is at \( x = -\frac{b}{2a} \). For our function with \( a = 1 \), \( b = -6 \), we calculate \( x = -\frac{-6}{2\times1} = 3 \). Substitute \( x = 3 \) in \( f(x) \) to find \( y \): \( f(3) = 3^2 - 6\times3 = 9 - 18 = -9 \). The vertex is \( (3, -9) \).

Step 3: Identify the X-Intercepts

The \( x \)-intercepts are found by setting \( f(x) = 0 \). Solve \( x^2 - 6x = 0 \) by factoring: \( x(x - 6) = 0 \). Thus, \( x = 0 \) and \( x = 6 \) are the \( x \)-intercepts.

Step 4: Find the Y-Intercept

The \( y \)-intercept is found by evaluating \( f(x) \) at \( x = 0 \). This gives \( f(0) = 0^2 - 6\times0 = 0 \). Therefore, the \( y \)-intercept is \( (0, 0) \).

Step 5: Sketch the Graph

Plot the vertex \( (3, -9) \), the intercepts \( (0, 0) \) and \( (6, 0) \), and draw a parabola opening upwards, passing through these points. The axis of symmetry is the vertical line \( x = 3 \).

Key Concepts

Standard Form

Quadratic functions expressed in the standard form are a key building block in understanding their behavior and characteristics. The standard form of a quadratic function is given by \( f(x) = ax^2 + bx + c \), where:

• \( a \) determines the opening direction and the width of the parabola.
• \( b \) affects the position of the vertex.
• \( c \) represents the \( y \)-intercept of the parabola.

For example, the function \( f(x) = x^2 - 6x \) is already in standard form. Here, the coefficients are \( a = 1 \), \( b = -6 \), and \( c = 0 \). This makes it simple to apply other algebraic methods to find more features of the graph. The coefficient \( a = 1 \) indicates that the parabola opens upwards and is relatively narrow.

Vertex

The vertex of a quadratic function provides us with the maximum or minimum point of the graph, depending on whether the parabola opens upwards or downwards. The vertex can be found using the formula:\[x = -\frac{b}{2a}\]After finding \( x \), substitute it back into the function to find the \( y \)-value of the vertex.
For example, in the function \( f(x) = x^2 - 6x \), the value for \( x \) is calculated as:\[x = -\frac{-6}{2 \times 1} = 3\]Substitute \( x = 3 \) into the original function to find the \( y \)-value:\[f(3) = 3^2 - 6 \times 3 = 9 - 18 = -9\]Thus, the vertex of this function is at the point \((3, -9)\). This tells us the lowest point (minimum) on the graph because the parabola opens upwards.

Intercepts

Intercepts are crucial points where the graph intersects the axes and they provide insightful information about the quadratic function. There are two types of intercepts:

• X-intercepts: Points where the graph crosses the \(x\)-axis. Found by solving the equation \( f(x) = 0 \).
• Y-intercept: The point where the graph crosses the \(y\)-axis. Found by evaluating the function at \( x = 0 \).

For the function \( f(x) = x^2 - 6x \), we calculate:

• The \( x \)-intercepts are found by solving the equation \( x^2 - 6x = 0 \), which factors to \( x(x - 6) = 0 \). Thus, the x-intercepts are \( x = 0 \) and \( x = 6 \).
• The \( y \)-intercept is simply \( c = 0 \), so the \( y \)-intercept is at \( (0, 0) \).

Parabola Sketching

Sketching a parabola involves plotting significant points such as the vertex and intercepts and understanding the parabola's symmetry and direction.Here's a simple way to sketch the parabola:

• Plot the vertex, which in our example is \((3, -9)\).
• Mark the \( x \)-intercepts at \((0, 0)\) and \((6, 0)\).
• Indicate the \( y \)-intercept located at \((0, 0)\).
• Draw the parabola opening upwards through these points, creating the U-shape characteristic of quadratic functions.
• The axis of symmetry passes through the vertex, providing a mirror line, which is \( x = 3 \) in this example.

These steps form the basis of graphing any quadratic function, ensuring that the major points and the symmetry of the parabola are accurately represented.