Problem 30 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$f(x)=x^{2}+3 x+\frac{1}{4}$$

Step-by-Step Solution

Answer

The vertex of the parabola is \(-\frac{3}{2}, f(-\frac{3}{2})\). Use the quadratic formula to find the x-intercepts. Then, sketch the graph of the function indicating the vertex and the x-intercepts.

Step 1: Identify Form

Recognize that this quadratic function is in the form \(f(x) = ax^{2} + bx + c\), where \(a = 1\), \(b = 3\), and \(c = \frac{1}{4}\).

Step 2: Find the Vertex

Use the formula for the x-coordinate of the vertex of a parabola, \(x = -\frac{b}{2a}\). Substituting \(b = 3\) and \(a = 1\), we get \(x = -\frac{3}{2}\). Substitute \(x = -\frac{3}{2}\) into the function to get the y-coordinate of the vertex. Thus, the vertex of the parabola is \(-\frac{3}{2}, f(-\frac{3}{2})\).

Step 3: Find the X-intercepts

To find the x-intercepts, set \(f(x) = 0\) and solve for \(x\). So the equation to solve is \(x^{2} + 3x + \frac{1}{4} = 0\). Use the quadratic formula to find the solutions.

Step 4: Sketch the Graph

Now, with the vertex and x-intercepts, plot these points on a graph. Since the leading coefficient \(a = 1\) which is greater than zero, the parabola opens upwards. Graph the parabola which passes through the points identified, reaching a minimum at the vertex point.

Key Concepts

Vertex of a Parabola

The vertex of a parabola is an important point that represents the ‘turning point’ of its graph. It is the highest or lowest point on the graph depending on whether the parabola opens downwards or upwards. In a quadratic function written in the standard form, \(f(x) = ax^2 + bx + c\), the x-coordinate of the vertex can be found using the formula: \(x = -\frac{b}{2a}\).
For the function \(f(x) = x^2 + 3x + \frac{1}{4}\), substitute \(a = 1\) and \(b = 3\) into the formula, which gives:

• \(x = -\frac{3}{2}\)

Next, calculate the y-coordinate by substituting \(x\) back into the function, \(f\left( -\frac{3}{2} \right)\). This substitution finds the value of \(f(x)\) at \(-\frac{3}{2}\). Thus, the vertex of this parabola is at \(\left(-\frac{3}{2}, f\left(-\frac{3}{2}\right)\right)\). This indicates the minimum point because the parabola opens upwards (since \(a = 1\) is positive).
This signifies that the y-coordinate at this x-value will be the smallest y-value on the graph of this parabola.

X-intercepts

X-intercepts, also known as roots or zeros of the function, are the points where the graph of the function crosses the x-axis. At these points, the value of the function is zero, hence \(f(x) = 0\). To find the x-intercepts for the quadratic equation \(x^2 + 3x + \frac{1}{4} = 0\), we can use the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(a = 1\), \(b = 3\), and \(c = \frac{1}{4}\). Plug these values into the formula to discover the x-values where the parabola intersects the x-axis.After computation, the quadratic formula might result in solutions that are not integers; therefore, it’s essential to simplify the expression carefully and maintain accuracy. These solutions provide the exact points where the parabola will touch or intersect the x-axis, indicating the real roots of the quadratic equation.

Graph of a Parabola

Visualizing the graph of a parabola helps in understanding its behavior and all related features. The given quadratic function \(f(x) = x^2 + 3x + \frac{1}{4}\) creates a parabola that opens upwards because the coefficient of \(x^2\), which is \(a = 1\), is positive.
To sketch the parabola:

• Plot the vertex, which is the lowest point of the parabola, found at \(-\frac{3}{2}, f\left(-\frac{3}{2}\right)\).
• Identify and plot the x-intercepts using the solutions from the quadratic formula.
• Draw a smooth curve to pass through the vertex and intercepts.

The shape is symmetric around a vertical line called the axis of symmetry, which passes through the vertex. The symmetry ensures that the points to the left and right of this line mirror each other. Given the calculations, the precise shape and position of each part of the parabola are determined.
This clear visual representation on the graph not only confirms the calculated vertex and x-intercepts but also allows for the visualization of how the parabola behaves further away from these key points, showing how it extends infinitely upwards.