Problem 16 - Algebra and Trigonometry

Question

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. $$ f(x)=-x^{2}-4 x+4 $$

Step-by-Step Solution

Answer

The function is already in standard form: \( f(x) = -x^2 - 4x + 4 \). The vertex is \((2, -8)\), x-intercepts are \(x = -2\sqrt{2}\) and \(x = 2 + 2\sqrt{2}\), y-intercept is \((0, 4)\). The graph is a downward-opening parabola.

Step 1: Express in Standard Form

The standard form of a quadratic function is given by \( ax^2 + bx + c \). The given function \( f(x) = -x^2 - 4x + 4 \) is already in standard form with \( a = -1 \), \( b = -4 \), and \( c = 4 \).

Step 2: Find the Vertex

The vertex of a quadratic function \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). For our function, \( a = -1 \) and \( b = -4 \), so \( x = -\frac{-4}{2(-1)} = 2 \). Plug this back into the function to find \( y \): \( f(2) = -2^2 - 4 \times 2 + 4 = -4 - 8 + 4 = -8 \). Therefore, the vertex is \((2, -8)\).

Step 3: Find the x-intercept(s)

To find the x-intercepts, set \( f(x) = 0 \): \(-x^2 - 4x + 4 = 0\). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), solve for \( x \): \( x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-1)(4)}}{2(-1)} = \frac{4 \pm \sqrt{16 + 16}}{-2} = \frac{4 \pm \sqrt{32}}{-2} = \frac{4 \pm 4\sqrt{2}}{-2} \). Thus, the x-intercepts are \( x = -2\sqrt{2} \, or \, x = 2 + 2\sqrt{2} \).

Step 4: Find the y-intercept

The y-intercept of a quadratic is found by evaluating \( f(0) \). Substitute \( x = 0 \) into the function: \( f(0) = -(0)^2 - 4(0) + 4 = 4 \). Thus, the y-intercept is \((0, 4)\).

Step 5: Sketch the Graph

With the information from previous steps, sketch the function. The parabola opens downwards (since \( a = -1 < 0 \)), with vertex at \((2, -8)\), x-intercepts at \(x = -2\sqrt{2}\) and \(x = 2 + 2\sqrt{2}\), and y-intercept at \((0, 4)\). Draw a symmetric curve around the vertex.

Key Concepts

Vertex

The vertex of a quadratic function is a crucial point on its graph. It represents the highest or lowest point, depending on the direction the parabola opens. To find the vertex of a quadratic function given in standard form \( f(x) = ax^2 + bx + c \), we use the formula for the x-value of the vertex:

• \( x = -\frac{b}{2a} \)

After finding this x-value, substitute it back into the function \( f(x) \) to get the y-coordinate. This gives you the vertex as the point \((x, y)\). For the function \( f(x) = -x^2 - 4x + 4 \), the vertex calculation goes as follows:

• First, find \( x = -\frac{-4}{2(-1)} = 2 \)
• Next, find \( y \) by substituting \( x = 2 \) into the function to get \( f(2) = -8 \)
• The vertex is \((2, -8)\)

In this case, because \( a = -1 \), the parabola opens downwards, and the vertex is the maximum point on the graph. The vertex helps understand the shape and location of the parabola.

x-intercepts

The x-intercepts of a quadratic function are points where the graph crosses the x-axis. These points occur when the output \( f(x) = 0 \). For finding x-intercepts, we solve the quadratic equation \( ax^2 + bx + c = 0 \). One of the most common methods is using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

For \( f(x) = -x^2 - 4x + 4 \), applying this formula gives:

• \( x = \frac{4 \pm \sqrt{32}}{-2} = \frac{4 \pm 4\sqrt{2}}{-2} \)
• Which simplifies to \( x = -2\sqrt{2} \) or \( x = 2 + 2\sqrt{2} \)

These values are the x-intercepts of the parabola. They tell us where the function equals zero and where the graph crosses the x-axis. Knowing the x-intercepts helps plot the points that are critical for graphing quadratic functions.

y-intercept

The y-intercept of a quadratic function is the point where the graph crosses the y-axis. This occurs when the input \( x = 0 \). To find the y-intercept, simply evaluate the function at \( x = 0 \). For the quadratic function \( f(x) = -x^2 - 4x + 4 \), plugging in \( x = 0 \) gives:

• \( f(0) = 4 \)
• Thus, the y-intercept is the point \((0, 4)\)

This intercept is essential for graphing, as it’s a quick point to identify on the graph. The y-intercept also provides a vertical reference point from which you can start sketching the parabola.

Standard Form

The standard form of a quadratic function is a specific way of writing the function, expressed as \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. The standard form is often preferred because it clearly reveals the coefficients that will be used in finding the vertex, x-intercepts, and y-intercept. For the quadratic \( f(x) = -x^2 - 4x + 4 \), it’s already in standard form:

• \( a = -1 \)
• \( b = -4 \)
• \( c = 4 \)

This form also helps determine the direction the parabola opens, based on the sign of \( a \). If \( a > 0 \), the parabola opens upwards; if \( a < 0 \), like in our function, it opens downwards. Understanding the standard form is crucial for working with and analyzing quadratic functions.