Problem 16 - Beginning and Intermediate Algebra

Question

For quadratic function, identify the vertex, axis of symmetry, and \(x\)- and \(y\)-intercepts. Then, graph the function. \(f(x)=-(x-2)^{2}-4\)

Step-by-Step Solution

Answer

The vertex of the quadratic function \(f(x)=-(x-2)^2-4\) is \((2, -4)\), and the axis of symmetry is given by the line \(x=2\). There are no real x-intercepts, and the y-intercept is \((0, -8)\). The graph shows a downward-opening parabola with its vertex at \((2, -4)\), symmetric with respect to the line \(x=2\), and passing through the y-intercept at \((0, -8)\).

Step 1: Find the vertex

The vertex of a quadratic function in the form \(f(x) = a(x - h)^2 + k\) is given by the point \((h, k)\). In this case, the function is already in vertex form: \(f(x) = -(x - 2)^2 - 4\). Thus, the vertex is \((2, -4)\).

Step 2: Determine the axis of symmetry

The axis of symmetry of a quadratic function in vertex form passes through the vertex and is a vertical line that can be represented by the equation \(x = h\). In this case, the vertex has an x-coordinate of \(2\), so the axis of symmetry is the line \(x = 2\).

Step 3: Calculate the x-intercepts

To get the x-intercepts, we need to find the values of \(x\) where the function equals zero: \(f(x) = 0\). So, for our function \(-(x - 2)^2 - 4\), set the equation equal to zero. \(-(x - 2)^2 - 4 = 0\) Add 4 to both sides: \(-(x - 2)^2 = 4\) Divide both sides by \(-1\): \((x - 2)^2 = -4\) Since the right-hand side is negative, there are no real solutions for \(x\), meaning that there are no x-intercepts for this quadratic function.

Step 4: Calculate the y-intercept

To find the y-intercept, we need to determine the value of the function when \(x = 0\). So, we will substitute 0 into the function: \(f(0) = - (0 - 2)^2 - 4\) \(f(0) = -(-2)^2 - 4\) \(f(0) = - (4) - 4\) \(f(0) = - 8\) Thus, the y-intercept is the point \((0, -8)\).

Step 5: Graph the function

Using the information obtained from the previous steps, we can now graph the function: 1. The vertex is \((2, -4)\). 2. The axis of symmetry is the vertical line at \(x = 2\). 3. There are no x-intercepts. 4. The y-intercept is the point \((0, -8)\). With these key points, we can sketch the graph. The function will be a downward-opening parabola with its vertex at \((2, -4)\), passing through the y-intercept at \((0, -8)\), and symmetric with respect to the vertical line \(x = 2\).

Key Concepts

Vertex Form

Understanding the vertex form of a quadratic function is crucial for easily identifying the vertex of the parabola it describes. In any quadratic expressed as \(f(x) = a(x - h)^2 + k\), the vertex can be directly extracted from \((h, k)\). This form shows us:

• The horizontal shift, represented by \(h\). If \(h > 0\), the graph shifts to the right, and if \(h < 0\), it shifts to the left.
• The vertical shift, represented by \(k\). If \(k > 0\), the graph moves up, and if \(k < 0\), it moves down.
• The coefficient \(a\), which affects the "width" and the "direction" (upwards if \(a > 0\), downwards if \(a < 0\)) of the parabola.

For the function \(f(x) = -(x-2)^2 - 4\), the vertex is at \((2, -4)\). This tells us the graph is horizontally shifted 2 units to the right and 4 units down from the origin, and is "upside down" due to the negative \(a\).

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. Perfect symmetry is one of the fundamental properties of a parabola, and it can be easily identified for quadratic functions expressed in vertex form. This line always passes through the vertex, and its equation is given by \(x = h\), where \(h\) represents the x-coordinate of the vertex. In the example function \(f(x) = -(x - 2)^2 - 4\), the axis of symmetry is \(x = 2\). This means that any point on the parabola at a certain distance to the left of \(x = 2\) will have a corresponding point at the same distance to its right, creating a perfect reflection across the line \(x = 2\).

X-Intercepts

The x-intercepts are the points where the graph of the function crosses the x-axis. These intercepts occur where the function's output \(f(x) = 0\). Finding these helps understand where the function changes sign from positive to negative or vice versa. Let's solve for our example:

• Given the equation \(-(x-2)^2 - 4 = 0\), we first add 4 to both sides to get \(-(x-2)^2 = 4\).
• Next, divide by -1: \((x-2)^2 = -4\).
• Since a square is never negative, there are no real solutions. Hence, the function \(f(x) = -(x-2)^2 - 4 \) has no x-intercepts.

In other words, this parabola doesn't cross the x-axis, staying in the negative y-region regardless of x values.

Y-Intercepts

Y-intercepts occur where the function crosses the y-axis, which happens when the input \(x = 0\). Finding the y-intercept is straightforward: substitute zero in place of \(x\) and evaluate the function. Let's apply this to our function:

• Start with \(f(0) = -(0-2)^2 - 4\).
• Calculate the inside first: \(-(2)^2 - 4 = -4 - 4\).
• This simplifies to \(f(0) = -8\).

Therefore, the y-intercept of \(f(x) = -(x-2)^2 - 4\) is at \((0, -8)\). This shows where our parabola touches the y-axis as it opens downward centered around its vertex.