Problem 30 - A Graphical Approach to Precalculus with Limits

Question

For quadratic function, (a) use the vertex formula to find the coordinates of the vertex and (b) graph the function. Do not use a calculator. $$P(x)=2 x^{2}-8 x+9$$

Step-by-Step Solution

Answer

The vertex is at (2, 1), and the graph is an upward-opening parabola.

Step 1: Identify the coefficients

The quadratic function given is \(P(x) = 2x^2 - 8x + 9\). Here, the coefficient \(a\) is 2, \(b\) is -8, and \(c\) is 9.

Step 2: Apply the vertex formula for the x-coordinate

The x-coordinate of the vertex for a quadratic function \(ax^2 + bx + c\) is given by \(x = \frac{-b}{2a}\). Substituting the values, we have \(x = \frac{-(-8)}{2(2)} = \frac{8}{4} = 2\).

Step 3: Calculate the y-coordinate of the vertex

Substitute \(x = 2\) into the original quadratic function to find the y-coordinate: \(P(2) = 2(2)^2 - 8(2) + 9 = 8 - 16 + 9 = 1\). Thus, the y-coordinate is 1.

Step 4: Determine the vertex coordinates

The vertex of the quadratic function \(P(x) = 2x^2 - 8x + 9\) is at the point \((2, 1)\).

Step 5: Graph the quadratic function

Since \(a = 2 > 0\), the parabola opens upwards. The vertex \((2, 1)\) is the lowest point on the graph. Find additional points such as \(P(0) = 9\) and \(P(1) = 3\) to help sketch the graph. Plot these points and draw the direction of the parabola opening upwards.

Key Concepts

Quadratic Function

A quadratic function is an essential concept in algebra, characterized by its unique polynomial form. It takes the expression form of \(ax^2 + bx + c\). In this structure:

• \(a\), \(b\), and \(c\) are constants where \(a eq 0\).
• \(a\) represents the coefficient of the quadratic term \(x^2\).
• \(b\) is the coefficient of the linear term \(x\).
• \(c\) is the constant term, also known as the y-intercept when the function is graphed.

In the exercise, the given quadratic function is \(P(x) = 2x^2 - 8x + 9\). The vertex formula, \(x = \frac{-b}{2a}\), helps us find the x-coordinate of the vertex. Finding the vertex is an integral aspect of understanding and visualizing the graph of a quadratic function.

Parabola

The graph of any quadratic function is a parabola. These U-shaped curves can open either upward or downward based on the coefficient \(a\). Here's how this works:

• If \(a > 0\), the parabola opens upwards, forming a cup-like shape.
• If \(a < 0\), the parabola opens downwards, resembling an upside-down cup.

In our case, where \(P(x) = 2x^2 - 8x + 9\), \(a\) is 2, meaning the parabola opens upwards. This ensures that the vertex of the parabola is its lowest point, minimizing the value of the function. The vertex calculated as \((2, 1)\) tells us not just the direction but also the base location on a graph. Understanding the properties of a parabola is crucial as it finds application in physics, engineering, and various other real-world phenomena.

Graphing

Graphing a quadratic function involves plotting points to sketch its U-shaped form accurately. Start by identifying key points: the vertex and additional points for reference. For the equation \(P(x) = 2x^2 - 8x + 9\):

• The vertex is \((2, 1)\).
• Substitute different x-values, such as \(x = 0\) or \(x = 1\), to find corresponding y-values like \(P(0) = 9\) and \(P(1) = 3\).
• Plot these points on the Cartesian plane.

As \(a > 0\), ensure that your sketch opens upward from the vertex. Because the function graph is symmetrical about a vertical line through the vertex, this line, known as the axis of symmetry, aids in perfectly balancing your parabola on both sides of it. Once all points are marked, sketch a smooth curve (the parabola) through them. This graph offers invaluable insight, not just visually, but also helps in understanding how changing the equation's coefficients affect its shape and direction.