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

Step-by-Step Solution

Answer

The vertex is (3, 2), the axis of symmetry is \(x = 3\), the parabola has no x-intercepts and the y-intercept is (0, 11), the domain is \(-\infty, \infty\) and the range is \([2, \infty)\).

Step 1: Identify the vertex

For the function \(f(x) = (x-3)^2 + 2\), the vertex is given by \((h, k)\) where h is 3 and k is 2. So, the vertex of the parabola is \((3, 2)\).

Step 2: Find the axis of symmetry

The axis of symmetry is determined by the equation \(x = h\). Here, \(h = 3\), so the axis of symmetry is \(x = 3\).

Step 3: Identify the intercepts

The x-intercepts are the solutions to the equation \(0 = (x-3)^2 + 2\). Solving this gives complex roots, indicating that the parabola does not intersect the x-axis. The y-intercept is obtained by evaluating the function at \(x = 0\). This gives \(f(0) = (0-3)^2 + 2 = 11\). So, the y-intercept is (0, 11).

Step 4: Determine the domain and range

The domain of a quadratic function is all real numbers (\(-\infty, \infty\)), since the x-value can be any real number. The range is determined by the y-value of the vertex and whether the parabola opens upwards or downwards. In this case, the parabola opens upwards and the y-value of the vertex is 2. So, the range is \([2, \infty)\).

Key Concepts

Vertex of a parabola

The vertex of a parabola is a key feature that gives you valuable information about the graph of a quadratic function. In the standard form of a quadratic function, \(f(x) = a(x-h)^2 + k\), the vertex is the point \((h, k)\). It represents the highest or lowest point on the graph, depending on whether the parabola opens upwards or downwards.

• If the parabola opens upwards, the vertex is a minimum point.
• If it opens downwards, the vertex is a maximum point.

In our example, \(f(x) = (x-3)^2 + 2\), the vertex is at the point \((3, 2)\). This location tells us that the parabola reaches its minimum value of 2 at this point. The vertex is a great tool for easily sketching the basic shape of a quadratic function.

Axis of symmetry

The axis of symmetry of a parabola is a vertical line that intersects the parabola at its vertex, dividing it into two symmetrical halves. This line provides an important guideline when sketching the graph of a quadratic function. In the equation \(f(x) = a(x-h)^2 + k\), the axis of symmetry is given by the equation \(x = h\).

• This means the axis of symmetry is the same as the x-coordinate of the vertex.
• It helps to know that any point on the parabola to the left of this line will have a mirror image on the right side.

For our example function, \(f(x) = (x-3)^2 + 2\), the axis of symmetry is \(x = 3\). Understanding this line's role makes it easier to visualize the symmetry of the parabola.

Domain and range of a function

The domain of a quadratic function is all real numbers, because there are no restrictions on the values that \(x\) can take. This means you can plug any real number into the function, and you'll get a resulting \(y\) value.

The range of a quadratic function, however, is determined by the vertex and the direction in which the parabola opens. For parabolas that open upwards, the range starts at the y-coordinate of the vertex and extends upwards to infinity.

• For upwards opening parabolas, the range is \([k, \infty)\), where \(k\) is the y-coordinate of the vertex.
• For downward opening parabolas, the range is \((-\infty, k]\).

In our exercise, the parabola described by \(f(x) = (x-3)^2 + 2\) opens upwards, with the vertex at \((3, 2)\). Thus, the range of this function is \([2, \infty)\). Understanding domain and range helps neatly define where the function exists and how far it extends.

Parabola intercepts

Intercepts are the points where a graph crosses the x-axis or y-axis. These points are important for understanding the position and shape of the parabola on the coordinate plane.

The x-intercepts are found by setting the function equal to zero and solving for \(x\).

• Not all parabolas will have x-intercepts; this happens when the graph doesn't cross the x-axis.

For the function \(f(x) = (x-3)^2 + 2\), solving \(0 = (x-3)^2 + 2\) reveals that there are no real x-intercepts, only complex solutions. This means the parabola does not cross the x-axis.

The y-intercept is found by evaluating the function at \(x = 0\). For our function, \(f(0) = (0-3)^2 + 2 = 11\), so the y-intercept is \((0, 11)\). This information helps to anchor the graph and understand its behavior in relation to the axes.