Problem 18 - College Algebra

Question

Use the vertex and intercepts to sketch the graph of each quadratic function. Give the equation of the parabola's axis of symmetry. Use the graph to determine the function's domain and range. \(f(x)=(x-1)^{2}-2\)

Step-by-Step Solution

Answer

The vertex of the function \(f(x) = (x - 1)^{2} - 2\) is \((1,-2)\), giving an axis of symmetry at \(x = 1\). The x-intercepts are \(x = \sqrt{2} + 1\) and \(x = -\sqrt{2} + 1\) and the y-intercept is \(-1\). The domain is all real numbers, and the range is \(f(x) \ge -2\).

Step 1: Identify the Vertex

The vertex form of a quadratic function is \(f(x) = a(x-h)^{2} + k\), where \((h, k)\) are the coordinates of the vertex. Here, the quadratic function is \(f(x) = (x-1)^{2} -2\), so the vertex is at \((h, k) = (1, -2)\).

Step 2: Find the Axis of Symmetry

The axis of symmetry of a quadratic function in vertex form is \(x = h\). So for the function \(f(x) = (x-1)^{2} - 2\), the axis of symmetry is \(x = 1\).

Step 3: Find the Intercepts

The x-intercepts are found by setting \(f(x) = 0\) and solving for \(x\). So solving the equation \((x-1)^{2} - 2 = 0\), we get \(x = \sqrt{2} + 1\) and \(x = -\sqrt{2} + 1\). The y-intercept is found by evaluating \(f(0)\), giving an intercept of \(f(0) = -1\).

Step 4: Sketch the Graph

Plot the vertex, the axis of symmetry, and the intercepts on a graph. As the parabola opens upwards, the vertex is the lowest point on the graph.

Step 5: Find the Domain and Range

For any quadratic function, the domain is all real numbers because the function is defined for each value of \(x\). The range is all values of \(f(x)\) that are equal to or higher than the y-coordinate of the vertex because the graph opens upwards. Therefore, in this case, the range is \(f(x) \ge -2\).

Key Concepts

Vertex of a Parabola

The vertex of a parabola is one of the most critical aspects when sketching its graph. Think of the vertex as the 'turning point' that determines the shape and orientation of the parabola. In our exercise with the quadratic function \(f(x)=(x-1)^{2}-2\), the vertex is found using the form \(f(x) = a(x-h)^{2} + k\), where the vertex is the point \( (h, k) \). Here, we have \( h = 1 \) and \( k = -2\), which gives us the vertex \( (1, -2)\). Since the coefficient of \( (x-1)^2 \) is positive, the parabola opens upwards, making this vertex the minimum point of the graph.

In practical terms, when sketching the parabola, start by plotting the vertex on the coordinate grid. Always check the leading coefficient \( a \) of \( x \) squared term to decide if the parabola opens upwards (if \( a \) is positive) or downwards (if \( a \) is negative). For \( (x-1)^{2} -2 \) since \( a = 1 \) which is positive, the parabola opens upward.

Axis of Symmetry

The axis of symmetry in a quadratic function is a vertical line that divides the parabola into two mirror images. For the given quadratic function \(f(x) = (x-1)^{2} - 2\), this axis of symmetry can be determined from the vertex form of the equation. It’s the line defined by the equation \(x = h\), where again \(h\) is the x-coordinate of the vertex. Therefore, for our function, the axis of symmetry is \(x = 1\).

When sketching the graph, the axis of symmetry provides a helpful guide. Points on one side of this axis can be 'mirrored' to the other to ensure the parabola is accurately represented. This axis not only aids in graphing but also provides a deeper understanding of the function's behavior and its solutions' symmetry.

Intercepts of a Quadratic Equation

Intercepts are the points where the parabola crosses the x-axis and y-axis. The x-intercepts (or roots) occur where \(f(x) = 0\). For our example, setting \( (x-1)^{2} - 2 = 0 \) reveals the two x-intercepts: \(x = \sqrt{2} + 1\) and \(x = -\sqrt{2} + 1\). The y-intercept is where the parabola crosses the y-axis, which occurs when \(x = 0\). Evaluating \(f(0)\) gives us the y-intercept at \(f(0) = -1\).

These intercepts are essential when plotting the parabola on a graph — they provide concrete points to which the curve must conform. By plotting the x-intercepts and the y-intercept, you can sketch a more accurate representation of the quadratic function.

Domain and Range of a Function

The domain of a function includes all the possible input values (or \(x\)-values), whereas the range includes all possible output values (or \(f(x)\)-values). For quadratic functions, the domain is always all real numbers because there's no \(x\)-value for which the function doesn't provide an output. Therefore, for our function \(f(x) = (x-1)^{2} - 2\), the domain is all real numbers.

On the other hand, the range is determined by the 'direction' of the parabola and its vertex. Since the given parabola opens upwards, and the vertex is at \( (1, -2) \)—which is its lowest point—the range is all the values of \(f(x)\) that are greater than or equal to the vertex's \(y\)-coordinate. Hence, the range for this function is \( f(x) \ge -2 \). Understanding the domain and range is crucial as it describes the extent of the function's inputs and outputs.