Problem 19 - Algebra and Trigonometry

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 parabola is at (1,2) and the axis of symmetry is \(x = 1\). There are no x-intercepts and the y-intercept is at (0,3). The graph of the function opens upwards. The domain of the function is all real numbers and its range is all real numbers equal or greater than 2.

Step 1: Identify the Vertex of the Parabola

The given quadratic function is in vertex form \(f(x) = a(x - h)^2 + k\), where \((h, k)\) are the coordinates of the vertex. Here, \(a = 1\), \(h = 1\), and \(k = 2\). Thus, the vertex of the parabola is \((1, 2)\).

Step 2: Find the Axis of Symmetry

The axis of symmetry for a parabola in the vertex form is always \(x = h\). So, in this case, the axis of symmetry is \(x = 1\).

Step 3: Determine the Intercepts

To find the x-intercepts, set \(y=f(x)\) to zero and solve for \(x\). Here, however, it's clear that there are no x-intercepts because the whole of \(f(x)\) is always greater than 2. To find the y-intercept, set \(x = 0\). Here, \(f(0) = (0 - 1)^2 + 2 = 3\), implying that the y-intercept is at point \((0, 3)\).

Step 4: Sketch the Graph

Now, with the following points: the vertex at \((1,2)\), the y-intercept at \((0,3)\) and the axis of symmetry \(x=1\), sketch the graph. Because the leading coefficient \(a = 1\) is positive, the parabola opens upwards.

Step 5: Determine the Domain and Range

The domain of a quadratic function is all real numbers, because for any input \(x\), \(f(x)\) will provide a real number. The range will be all values equal to or above the y-coordinate of the vertex, because the parabola opens upwards. In this case, the range will be \(y \geq 2\).

Key Concepts

Vertex Form

The vertex form of a quadratic function is a specific way of expressing the quadratic equation that makes it easy to identify the vertex of the parabola directly from the equation itself. In the vertex form, a quadratic function is written as: \( f(x) = a(x - h)^2 + k \).

• \(a\) is a coefficient that determines the width and direction of the parabola.
• \(h\) and \(k\) are the coordinates of the vertex.

For example, in the function \( f(x) = (x - 1)^2 + 2 \), it's clear that the vertex is at the point \((1, 2)\). This form is especially useful as it immediately shows us where the parabola turns, making graphing or sketching quick and intuitive.

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. In a quadratic equation written in vertex form \( f(x) = a(x - h)^2 + k \), the axis of symmetry is easily identified as the line \( x = h \). This is because the vertex, located at \((h, k)\), is the peak or trough of the parabola. In our example, with the function \( f(x) = (x - 1)^2 + 2 \), the axis of symmetry this time is \( x = 1 \).

• The axis of symmetry always passes through the vertex of the parabola.
• Knowing this helps with graphing, ensuring both sides of the parabola reflect each other accurately.

Parabola

A parabola is the U-shaped graph you get from plotting a quadratic function. The orientation—whether it opens upwards or downwards—depends on the coefficient \( a \) in the quadratic equation.

• If \( a > 0 \), as in our example \( f(x) = (x - 1)^2 + 2 \), the parabola opens upwards.
• If \( a < 0 \), it opens downwards.

Parabolas have several crucial properties:

• They always have a vertex which is the highest or lowest point.
• The axis of symmetry plays a vital role in determining their symmetry.
• They may have x-intercepts where they cross the x-axis, but in this example, there are none because \( f(x) \) is always greater than 2.

Being able to sketch a parabola accurately depends on finding these features first.

Domain and Range

The domain and range of a quadratic function provide essential information about its behavior.

• The **domain** of any quadratic function is all real numbers. This means no matter what value of \( x \) you choose, the function \( f(x) \) will give you a valid y-value. Hence, for \( f(x) = (x - 1)^2 + 2 \), there are no restrictions on \( x \).

For the **range**, you need to look at the y-values that the function can output:

• Since the parabola opens upwards and the vertex's y-value is \( k \), the range is all values greater than or equal to \( k \).
• In our example, the smallest possible value of \( f(x) \) is 2, thus the range is \( y \geq 2 \).

Understanding the domain and range helps to ascertain the limits within which the function operates, making it easier to predict its graph's behavior.