Problem 8 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

For Problems 1 through 8, graph the function. Label the \(x\) - and \(y\) -intercepts and the coordinates of the vertex. $$ f(x)=x^{2}+\pi x+1 $$

Step-by-Step Solution

Answer

The x-intercepts, which are the roots of the function, can be found using the quadratic formula, producing complex roots. The y-intercept is at point (0, 1) and the x-coordinate of the vertex is \(-\pi / 2\). Substitute the x-coordinate of vertex in the equation to get its y-coordinate. Draw these points on a graph along with the indicative parabolic shape.

Step 1: Finding the x-intercepts

To find the x-intercepts, set \(f(x)\) equal to zero and solve for \(x\), like so: \(0 = x^{2} + \pi x + 1\). Unfortunately, this equation isn't solvable by basic factoring, so the quadratic formula will be needed: \(x = [-\pi \pm \sqrt{(\pi)^2 - 4(1)(1)}] / 2(1)\)

Step 2: Finding the y-intercept

The y-intercept of a function can be found by setting the variable \(x\) to zero: \(f(0) = 0^{2} + \pi x + 1 = 1\). Thus, the y-intercept is at point (0, 1)

Step 3: Finding the vertex of the function

For a quadratic function given in standard form, the x-coordinate of the vertex is given by \(-b/2a\): thus, vertex_x = \(-\pi / 2 * 1\). Substitute this value into the equation to find the \(y\) coordinate of the vertex: \(f(\mbox{vertex_x}) = (\mbox{vertex_x})^{2} + \pi * \mbox{vertex_x} + 1 \)

Step 4: Graphing the function

Based on the above results, plot the x and y intercepts and the vertex on a graph. Sketch the function, remembering that quadratic functions have a parabolic shape.

Key Concepts

Understanding X-Intercepts in Quadratic Functions

Quadratic functions can cross the x-axis at points known as x-intercepts, also referred to as roots or zeros. These points occur where the function's value equals zero, that is, where \(f(x) = 0\). For the function \(f(x) = x^2 + \pi x + 1\), we find x-intercepts by solving the equation \(x^2 + \pi x + 1 = 0\). Solving for x-intercepts often requires the use of the quadratic formula when the equation does not factor easily. The quadratic formula is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\). In our function, \(a = 1\), \(b = \pi\), and \(c = 1\). By substituting these coefficients into the formula, we get the roots, or x-intercepts. Remember, depending on the discriminant (\(b^2 - 4ac\)), our quadratic can have:

• Two distinct real roots
• One real root (if the discriminant is zero)
• No real roots (if the discriminant is negative)

In this case, we need to compute \(b^2 - 4ac\) to check the nature and real parts of the roots.

Determining the Y-Intercept

Finding the y-intercept of a quadratic function is a straightforward task. The y-intercept is the point where the graph of the function crosses the y-axis. This happens when the x-value is zero. Therefore, to find the y-intercept, we simply calculate \(f(0)\). For the function \(f(x) = x^2 + \pi x + 1\), substitute \(x = 0\): \[f(0) = 0^2 + \pi \, \times \, 0 + 1 = 1\]Thus, the y-intercept is at the point \((0, 1)\). When plotting a quadratic function like \(f(x) = x^2 + \pi x + 1\), the y-intercept provides a key point for shaping the graph. Remember, every quadratic function's graph is a parabola, which means it's symmetrical about its axis of symmetry.

Finding the Vertex of a Parabola

The vertex of a parabola is a crucial point that represents either the peak or the trough, depending on its orientation (upward or downward). The vertex lies on the axis of symmetry of the parabola. For a quadratic function in standard form \(f(x) = ax^2 + bx + c\), the x-coordinate of the vertex can be found using the formula:\[\text{vertex}_x = \frac{-b}{2a}\] For the function \(f(x) = x^2 + \pi x + 1\), use \(a = 1\) and \(b = \pi\) to calculate:\[\text{vertex}_x = \frac{-\pi}{2 \, \times \, 1} = -\frac{\pi}{2}\] To find the y-coordinate of the vertex, substitute \(\text{vertex}_x\) back into the function:\[f\left(-\frac{\pi}{2}\right) = \left(-\frac{\pi}{2}\right)^2 + \pi \, \times \, -\frac{\pi}{2} + 1\] By evaluating this, you will find the precise y-coordinate for the vertex. The vertex, in essence, is a point that helps give the parabola its distinct U-shaped graph. By knowing both the vertex and intercepts, you create a fuller picture of the function's graph.