Problem 16 - A Graphical Approach to College Algebra

Question

For each quadratic function defined , (a) write the function in the form \(P(x)=a(x-h)^{2}+k,\) (b) give the vertex of the parabola, and (c) graph the function. Do not use a calculator. $$P(x)=4 x^{2}+3 x-1$$

Step-by-Step Solution

Answer

(a) Vertex form: \(P(x) = 4(x + \frac{3}{8})^2 - \frac{25}{16}\). (b) Vertex: \((-\frac{3}{8}, -\frac{25}{16})\). (c) Parabola opens upward.

Step 1: Identify Components

The given quadratic function is \(P(x) = 4x^2 + 3x - 1\). In order to rewrite it in the form \(P(x)=a(x-h)^2+k\), identify the coefficients: \(a = 4\), \(b = 3\), and \(c = -1\).

Step 2: Complete the Square

To rewrite \(P(x)\) in vertex form, complete the square. Start with the terms involving \(x\): \[ P(x) = 4(x^2 + \frac{3}{4}x) - 1\]Take half of the coefficient of \(x\), \(\frac{3}{4}\), square it to get \(\left(\frac{3}{8}\right)^2 = \frac{9}{64}\), and adjust for maintaining equality:\[ P(x) = 4(x^2 + \frac{3}{4}x + \frac{9}{64} - \frac{9}{64}) - 1 \]This becomes:\[ P(x) = 4( (x + \frac{3}{8})^2 - \frac{9}{64}) - 1 \]

Step 3: Simplify the Expression

Distribute the \(4\) across the expression:\[ P(x) = 4(x + \frac{3}{8})^2 - 4\times\frac{9}{64} - 1 \]Simplify:\[ P(x) = 4(x + \frac{3}{8})^2 - \frac{9}{16} - 1 \]Combine constants:\[ P(x) = 4(x + \frac{3}{8})^2 - \frac{25}{16} \]

Step 4: Identify the Vertex

From the form \(P(x) = a(x-h)^2 + k\), the vertex \((h, k)\) is directly readable as the point:\[ h = -\frac{3}{8}, k = -\frac{25}{16} \]Thus, the vertex of the parabola is \((-\frac{3}{8}, -\frac{25}{16})\).

Step 5: Sketch the Graph

Graph the parabola using its vertex form \(P(x) = 4(x + \frac{3}{8})^2 - \frac{25}{16}\), with \(a = 4\), indicating it opens upward and is relatively narrow. Plot the vertex at \((-\frac{3}{8}, -\frac{25}{16})\), and draw the parabola, noting that the function is symmetric about the vertical line \(x = -\frac{3}{8}\).

Key Concepts

Completing the Square

Completing the square is a method to transform a quadratic equation into vertex form. The vertex form of a quadratic function is given by \(P(x) = a(x-h)^2 + k\). This format is ideal for identifying the vertex of the function, which describes the highest or lowest point of the parabola depending on the direction it opens.

The process involves a few key steps:

• First, identify and isolate terms with \(x\): For our example \(P(x) = 4x^2 + 3x - 1\), these are the \(4x^2\) and \(3x\).
• Next, prepare to complete the square by factoring out the coefficient of \(x^2\) from these terms, here it's \(4\): \(4(x^2 + \frac{3}{4}x)\).
• Find half of the \(x\) coefficient and square it. For \(\frac{3}{4}\), half is \(\frac{3}{8}\), and squaring it gives \(\frac{9}{64}\).
• Add and subtract the squared term inside the bracket to maintain balance: \(4(x^2 + \frac{3}{4}x + \frac{9}{64} - \frac{9}{64})\).

This step allows you to express \(P(x)\) as a perfect square trinomial plus or minus a constant, ultimately transforming it into vertex form.

Graphing Quadratics

Graphing quadratic functions involves plotting the U-shaped curve called a parabola. The shape and position of the parabola depend on its vertex form \(P(x) = a(x-h)^2 + k\). Using this form, you can quickly determine several important aspects of the graph.

Here's how you can approach graphing:

• Start with the vertex, \((h, k)\), which is the turning point of the parabola. In our exercise, the vertex is \((-\frac{3}{8}, -\frac{25}{16})\).
• Determine the direction the parabola opens based on the coefficient \(a\). If \(a > 0\), it opens upward; if \(a < 0\), it opens downward. Our example has \(a = 4\), so it opens upward, making it a standard concave-up parabola.
• Look at the axis of symmetry, which is the line \(x = h\). This line divides the parabola into two mirror-image halves. Here, it is \(x = -\frac{3}{8}\).
• Note the rate/width of curvature based on \(a\). A larger \(|a|\) value means a narrower parabola. Since \(a = 4\), the graph is relatively narrow.

Plot these elements on a graph, using the vertex and symmetry to guide the shape.

Properties of Parabolas

Parabolas, which are defined by quadratic functions, have some distinctive properties that influence their graph and behavior. Understanding these properties helps in analyzing and sketching parabolas effectively.

Key properties include:

• Vertex: The vertex \((h, k)\) is the highest or lowest point and is crucial for determining the parabola's maximum or minimum.
• Axis of Symmetry: A vertical line through \(x = h\) that splits the parabola into two equal halves. It's helpful for ensuring accuracy when sketching.
• Direction: Determined by the sign of \(a\), affecting if the parabola opens upwards or downwards.
• Y-intercept: Although not part of vertex form, it's found by setting \(x = 0\) and computing \(P(0)\). For \(P(x) = 4x^2 + 3x - 1\), it's simple to compute.
• Domain and Range: The domain for all quadratics is \(\mathbb{R}\), but the range depends on the direction and vertex. Range starts from \(k\) and goes upwards (if \(a > 0\)) or downwards (if \(a < 0\)). For our function, the range is \([-\frac{25}{16}, \infty)\).

These aspects together form the blueprint of a parabola, making it easier to understand and draw.