Problem 26 - Beginning and Intermediate Algebra

Question

For quadratic function, identify the vertex, axis of symmetry, and \(x\)- and \(y\)-intercepts. Then, graph the function. \(f(x)=2(x-3)^{2}+3\)

Step-by-Step Solution

Answer

Vertex: (3, 3) Axis of symmetry: x = 3 x-intercepts: None y-intercept: (0, 21) To graph the function, plot the vertex (3, 3), the y-intercept (0, 21), the axis of symmetry at x=3, and connect the points to form a parabola with symmetry around the axis of symmetry.

Step 1: Identify Vertex #

The given quadratic function is \(f(x) = 2(x-3)^2 +3\). Since the function is in vertex form \((f(x)=a(x-h)^2+k)\), we can directly read the vertex's coordinates \((h,k)\) from the equation. In our case, the vertex is point (3, 3).

Step 2: Find Axis of Symmetry #

The axis of symmetry of a parabola always passes through the vertex, and for a vertical parabola, it is a vertical line. The equation for the axis of symmetry in such cases will be x=h, with h being the x-coordinate of the vertex. Here, the axis of symmetry is x=3.

Step 3: Find x-intercepts #

To find the x-intercepts of the given function, we need to find the x-values when f(x) = 0: \(0 = 2(x-3)^2 + 3\) Now we solve for x: \(-3 = 2(x-3)^2\) \(\frac{-3}{2} = (x-3)^2\) Since the expression on the right-hand side of the equation is a squared term, it must be non-negative. However, the left-hand side is negative (-3/2), which means there are no x-intercepts in this particular function.

Step 4: Find y-intercepts #

To find the y-intercepts of the given function, we need to find the f(x) value when x = 0: \(f(0) = 2 (0-3)^2 + 3\) \(f(0) = 2(9) + 3\) \(f(0) = 18 + 3\) \(f(0) = 21\) So, the y-intercept of the function is the point (0, 21).

Step 5: Graph the Function #

To graph the function, follow these steps: 1. Draw the axis of symmetry (x=3). 2. Plot the vertex at point (3, 3). 3. Plot the y-intercept at point (0, 21). 4. Since we know that the function is symmetric with respect to the axis of symmetry (x=3), plot additional points by reflecting the y-intercept with respect to the axis of symmetry. 5. Draw the parabola by connecting the points plotted above, maintaining symmetry around the axis of symmetry (x=3). Now you have successfully graphed the given quadratic function: \(f(x)=2(x-3)^{2}+3\).

Key Concepts

Vertex Form

Quadratic functions can be expressed in a special format known as vertex form. This form is useful for quickly identifying certain characteristics of the parabola. The general equation is

• \(f(x) = a(x-h)^2 + k\)

In this formula:

• \(a\) controls the parabola's width and direction – wider, narrower, opening upwards or downwards.
• \(h\) and \(k\) determine the coordinates of the vertex \((h, k)\).

Vertex form is especially beneficial because it allows us to easily find the vertex, the point where the parabola either peaks or dips. From our equation \(f(x) = 2(x-3)^2 + 3\), the vertex is \((3, 3)\). Understanding the vertex form is key in graphing and analyzing quadratic functions.

Axis of Symmetry

The axis of symmetry is a vital part of understanding and graphing quadratic functions. This is a vertical line that divides the parabola into two mirror images. For parabolas in vertex form, the axis of symmetry can be quickly identified as:

• \(x = h\)

This line always passes through the vertex. In our function \(f(x) = 2(x-3)^2 + 3\), the axis of symmetry is at \(x = 3\).
It helps ensure that any point on the parabola on one side of the axis has a corresponding point on the other side, making graph plotting simpler and more accurate.

Graphing Parabolas

Graphing a parabola involves several steps, starting from identifying key features of the quadratic equation:

• Determine the vertex from the vertex form, as it gives a starting reference point.
• Locate the axis of symmetry which helps in ensuring the shape of the graph is accurate.
• Identify the intercepts to mark more specific points on the graph.
• Draw a smooth symmetric curve through these points, extending it as needed.

For example, with \(f(x) = 2(x-3)^2 + 3\), the graph is plotted as follows:

• Vertex at (3, 3).
• Axis of symmetry at \(x = 3\).
• Add points such as y-intercept at (0, 21).

Using symmetry around the axis forms the graph, ensuring the parabola is accurate and neatly plotted.

X-Intercepts

The x-intercepts of a quadratic function are points where the graph crosses the x-axis. This means that the function’s value is zero at these points, or:

• \(f(x) = 0\)

To find the x-intercepts from

• \(f(x) = 2(x-3)^2 + 3\)

we set the function to zero and solve:

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

This leads to a contradiction since squares are non-negative, implying no real roots. Therefore, for this function:

• there are no x-intercepts.

Understanding this helps in recognizing the function's behavior without crossing the x-axis.

Y-Intercepts

The y-intercepts are the points where the graph crosses the y-axis. This occurs when the x-value is zero. Finding the y-intercept involves substituting zero for x in the function:

• \(f(0) = 2(0-3)^2 + 3\)
• \(f(0) = 2 \cdot 9 + 3 = 21\)

Thus, the y-intercept is

• at the point (0, 21).

The y-intercept can give us a clear starting point for graphing when no other intercepts are available. This is especially useful in making a complete and balanced graph of a quadratic function.