Problem 20 - 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. $$ f(x)=(x-3)^{2}+2 $$

Step-by-Step Solution

Answer

The vertex of the given quadratic function \(f(x)=(x-3)^{2}+2\) is (3,2). The parabola does not have any x-intercepts. The equation of the axis of symmetry is \(x=3\). The domain is all real numbers. The range is all numbers y such that \(y \geq 2\).

Step 1: Determine the vertex and intercepts

The given function is in the vertex form \(f(x) = a(x-h)^2 + k\). Here, (h, k) = (3, 2) is the vertex of the parabola. To find the x-intercepts, set \(f(x) = 0\) and solve for x: \((x-3)^2 + 2 = 0\), that is \((x-3)^2= -2\). Since the square of a real number cannot be negative, there are no real solutions, so the graph does not intersect the x-axis.

Step 2: Determine the axis of symmetry

The axis of symmetry for a quadratic function in the form \(f(x) = a(x-h)^2 + k\) is given by x = h. Thus, for the given function, the axis of symmetry is \(x=3\). The parabola is symmetric about this vertical line.

Step 3: Determine the domain and range

For any quadratic function, the domain (possible x values) is all real numbers. The range (possible y values) depends on whether the parabola opens upwards or downwards and the y-coordinate of the vertex. Here, the parabola opens upwards (since the coefficient of \(x^2\) is positive), and the y-coordinate of the vertex is 2. Therefore, the range is all numbers y such that \(y \geq 2\).

Key Concepts

Vertex Form

Quadratic functions can be expressed in the vertex form, which is given by the equation \( f(x) = a(x-h)^2 + k \). This format makes it easy to identify the vertex of the parabola, which is a significant point on the graph. The vertex represents either the highest or lowest point of the parabola, depending on whether it opens upwards or downwards. In the present function \( f(x) = (x-3)^2 + 2 \), it is clear that:

• \( h = 3 \)
• \( k = 2 \)

Thus, the vertex of this parabola is \( (3, 2) \). This point is crucial because it gives us the "turning point" of the parabola, providing a clear image of its layout on the coordinate plane. In this particular case, since the vertex is positioned at \( (3, 2) \), it is at this point that the parabola changes direction.

Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex of the parabola and divides it into two mirror-image halves. For a quadratic function in vertex form \( f(x) = a(x-h)^2 + k \), the axis of symmetry can be found using the formula \( x = h \). In our function, \( f(x) = (x-3)^2 + 2 \), we see that the axis of symmetry is \( x = 3 \). This means the parabola is evenly balanced on this vertical line.

• All points on one side of the axis have corresponding points directly on the opposite side.
• The axis of symmetry is an integral part of sketching and understanding the graph, allowing one to determine the parabola’s shape and direction.

Domain and Range

Understanding the domain and range of a quadratic function is crucial for grasping the full scope of the function's graph. The domain of any quadratic function is all real numbers, meaning it can take any x-value from negative to positive infinity. This holds true for our function.For the range, one must observe the direction the parabola opens and the y-value of the vertex. Since our function \( f(x) = (x-3)^2 + 2 \) opens upwards (as indicated by the positive coefficient of the \( (x-3)^2 \) term), the range includes all values that are greater than or equal to the vertex's y-value. Specifically:

• The vertex is at \( y = 2 \), so the range is \( y \geq 2 \).
• No value of y below 2 is attainable from this function.

This range reflects that the parabola starts at the vertex and extends upwards indefinitely.

X-Intercepts

X-intercepts are points where the parabola crosses the x-axis, indicating when \( f(x) = 0 \). To find these intercepts, set the quadratic equation equal to zero and solve for x. With our function \( f(x) = (x-3)^2 + 2 \), solving for x when \( f(x) = 0 \) leads to:

• \((x-3)^2 + 2 = 0\)
• \((x-3)^2 = -2\)

As the square of any real number must be non-negative, there are no real solutions for \((x-3)^2 = -2\). Thus, our function has no x-intercepts, and it confirms that the parabola does not cross the x-axis. This can happen when the vertex is above or below the x-axis and the parabola opens upwards or downwards respectively.