Problem 21 - 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. $$y-1-(x-3)^{2}$$

Step-by-Step Solution

Answer

The vertex is at (3, 1), the x-intercept is (3, 0) and the y-intercept is (0, 10). The axis of symmetry is the line \( x = 3 \). The domain of the function is all real numbers, and the range is \( y ≥ 1 \).

Step 1: Identify Vertex and Sketch the Graph of Quadratic Function

Based on the standard form \( y = a(x - h)^{2} + k \) of a quadratic function, the vertex is at \((h, k)\). Here, the vertex is at \((3,1)\). Now, the parabola opens upwards because the coefficient of \( (x - h)^{2} \) is positive. So, graph it accordingly.

Step 2: Determine X-Intercepts

To find the x-intercepts, set \( y \) equal to 0 and solve for \( x \). Therefore, \( 0 = (x - 3)^{2} + 1 - 1 \), leading to \( (x - 3)^{2} = 0 \), and solving for \( x \) gives \( x = 3 \). Hence, the quadratic function intersects the x-axis at (3,0).

Step 3: Determine Y-Intercept

To find the y-intercept, set \( x \) equal to 0 in the equation. This gives us \( y = (0 - 3)^{2} + 1 = 10 \). Hence the y-intercept is at (0,10).

Step 4: Determine the Equation of the Axis of Symmetry

For a quadratic function, the axis of symmetry is the line \( x = h \) where \( h \) is the x-coordinate of the vertex. Therefore, the axis of symmetry for this function is \( x = 3 \).

Step 5: Determine the Domain and Range

The domain of a quadratic function is all real values of \( x \), because \( x \) can be any real number. The range is determined based on the direction the parabola opens. Since this parabola opens upwards and its vertex is at \( y = 1 \), the range is all real values of \( y \) such that \( y ≥ 1 \)

Key Concepts

Vertex

In the graph of a quadratic function, the vertex is a critical point that represents either the maximum or minimum of a parabola. In the equation provided, the quadratic function is in the form \( y = (x - h)^2 + k \). Here, the vertex is directly given by the coordinates \((h, k)\).

For the given function \( y = (x-3)^2 + 1 \), comparing this with the standard form reveals that the vertex is at \((3, 1)\). This point is crucial because it dictates the parabola's general shape and provides a point of symmetry.

On the graph, the vertex is where the curve stops moving downward (if it opens upwards) or upward (if it opens downwards), before it changes direction. Since the coefficient in front of \((x - h)^2\) is positive, the parabola opens upwards, confirming that our vertex \( (3, 1) \) is a minimum point.

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two mirror-image halves. For quadratic functions in the form \( y = a(x - h)^2 + k \), the axis of symmetry is always the line \( x = h \).

In our function \( y = (x-3)^2 + 1 \), the axis of symmetry is \( x = 3 \). This makes sense because the vertex, located at \( (3, 1) \), is positioned precisely on this line. This axis helps in drawing the parabola, particularly when trying to ensure it's symmetrical about this line.

Knowing the axis of symmetry is not only useful for graphing, but also for understanding properties of the function, like its vertex form and other transformations it might undergo.

Domain and Range

For any quadratic function, the domain is a set of all possible x values. Since parabolas extend indefinitely in the horizontal direction, the domain of any quadratic function is all real numbers, \(-\infty < x < \infty \).

However, the range depends on whether the parabola opens upwards or downwards. In our case, the function \( y = (x-3)^2 + 1 \) opens upwards based on the positive leading coefficient. Therefore, the range of the function starts from the vertex and extends upwards.

Since our vertex is at \((3, 1)\), the lowest point on the parabola is 1. Therefore, the range of the function is \( y \geq 1 \). This means that there are no y-values below 1 in the function's set of possible outputs.

Intercepts

Intercepts are the points where the graph intersects the x-axis or y-axis. They give significant clues about the graph's behavior.

• X-Intercept: Set \( y = 0 \) in the quadratic equation to find the x-intercept(s). For \( y = (x-3)^2 + 1 \), setting \( y \) to 0 leads to \((x-3)^2 = 0\), giving \( x = 3 \) as the x-intercept at (3, 0).
• Y-Intercept: Set \( x = 0 \) to find the y-intercept. Substituting \( x = 0 \) into \( y = (x-3)^2 + 1 \) yields \( y = 10 \). Thus, the y-intercept is (0, 10).

These intercepts help plot the graph by indicating where the parabola crosses the axes, aiding in visualizing shifts and transformations of the function.