Problem 29 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the quadratic function. Identify the vertex and intercepts. $$f(x)=x^{2}-x+\frac{5}{4}$$

Step-by-Step Solution

Answer

The vertex of the function is \(\left(\frac{1}{2},\frac{5}{4}\right)\), there are no x-intercepts, and the y-intercept is at y= \(\frac{5}{4}\). The graph of the function is a parabola that opens upwards, with its vertex at \(\left(\frac{1}{2},\frac{5}{4}\right)\), and touches the y-axis at y=\(\frac{5}{4}\).

Step 1: Vertex calculation

Calculate the vertex (h,k) of the quadratic function \(f(x)=x^{2}-x+\frac{5}{4}\). The x-coordinate of the vertex, h, is given by \(-b/2a\). Since a=1 and b=-1, the x-coordinate is \(1/2\). Substitute \(x=1/2\) into the equation to find the y-coordinate, k. Therefore, the vertex of the function is \(\left(\frac{1}{2},\frac{5}{4}\right)\).

Step 2: X-intercepts

Find the x-intercepts by setting the function equal to zero and solving for x. However, in this case, the function \(f(x)=x^{2}-x+\frac{5}{4}\) has no real solutions when set to zero, which means there are no x-intercepts.

Step 3: Y-intercept

Find the y-intercept by setting x=0 in the function. The y-intercept is \(f(0) = \frac{5}{4}\).

Step 4: Sketch the graph

Now with the vertex and intercepts, you're able to graph the quadratic function. Knowing that the function opens upwards (since a>0), you can plot the vertex at \(\left(\frac{1}{2},\frac{5}{4}\right)\) and know that the graph opens upwards from this point. Also knowing that there are no x-intercepts and the y-intercept is at y=\(\frac{5}{4}\), you can draw the parabola accordingly.

Key Concepts

Vertex of a Quadratic Function

The vertex of a quadratic function is the highest or lowest point of the graph, known as the maximum or minimum. In the function f(x) = ax^2 + bx + c, where a, b, and c are constants, the vertex can be found using the formula for the x-coordinate, which is h = -b/(2a). Once you have the x-coordinate, you can substitute it back into the function to find the y-coordinate, k.

In our exercise, the function f(x) = x^2 - x + 5/4, by applying the vertex formula, gives us h = 1/2. Substituting x = 1/2 into the function yields the y-coordinate, k = 5/4. Thus, the vertex of this function is at the point \(\left(\frac{1}{2}, \frac{5}{4}\right)\). Understanding the importance of the vertex is crucial as it is the turning point of the graph and indicates the direction in which the parabola opens.

X-intercepts and Y-intercepts

The points where a graph intersects the axes are known as intercepts. For a quadratic function, there can be zero, one, or two x-intercepts, which are the points where the graph crosses the horizontal axis. These intercepts are found by setting the quadratic equation equal to zero, f(x) = 0, and solving for x. In contrast, there is always a single y-intercept, where the graph crosses the vertical axis, which is found by evaluating the function at x = 0.

For the given function f(x) = x^2 - x + 5/4, no real solutions exist for x when the function is set to zero, indicating there are no x-intercepts. By substituting x = 0 in the equation, we find the y-intercept to be \(f(0) = \frac{5}{4}\). Intercepts provide critical points that help in sketching the overall graph of the quadratic function.

Graphing Parabolas

Graphing a parabola—the shape of a quadratic function's graph—requires identifying key features such as the vertex, axis of symmetry, intercepts, and the direction the parabola opens (upward or downward). To sketch a parabola, one must plot the vertex and intercepts on a coordinate plane, then use the shape's symmetry about the vertex to fill in the graph.

In our example with f(x) = x^2 - x + 5/4, we graph the vertex at \(\left(\frac{1}{2}, \frac{5}{4}\right)\), note the absence of x-intercepts, and place the y-intercept at \(0, \frac{5}{4}\). The parabola opens upwards since the coefficient a in f(x) = ax^2 + bx + c is positive. By understanding these key aspects and plotting several points around the vertex, specifically using the symmetry of the graph, students can accurately sketch the parabola's curve.