Problem 50 - Algebra 1

Question

SKETCHING GRAPHS Sketch the graph of the function. Label the vertex. $$ y=2 x^{2}-x $$

Step-by-Step Solution

Answer

The vertex of the function \(y = 2x^2 - x\) is \((1/4, -1/8)\), and the graph of this function is a parabola that opens upward with this vertex.

Step 1: Identify the Vertex

The first step is to identify the vertex. The vertex of a parabola in the form \(y = ax^2 + bx + c\) is \((-b/2a, (4ac-b^2)/4a)\). Here, \(a = 2\), and \(b = -1\). So, \(h = -b/(2a) = -(-1)/(2*2) = 1/4\). Then, substitute \(x = 1/4\) into the original function to calculate \(k = f(h)\), which gives \(k = 2 * (1/4)^2 - 1/4 = 2 * 1/16 - 1/4 = 1/8 - 1/4 = -1/8\). So, the vertex is \((1/4, -1/8)\).

Step 2: Sketch the Parabola

You need to plot the function. Start by plotting the vertex \((1/4, -1/8)\). Because a is positive, the parabola opens upward. Draw a curve that opens upward, with the vertex at the point plotted earlier. The y-intercept is found by setting x to 0, which gives \(y = 2 * 0^2 - 0 = 0\). So, plot the y-intercept \((0, 0)\). You could plot more points for a more accurate drawing. Finally, draw a smooth curve through the points, this is the graph of \(y = 2x^2 - x\).

Key Concepts

Vertex

The vertex of a parabola is a pivotal concept in understanding quadratic functions. The vertex is essentially the "turning point" of the parabola, representing the maximum or minimum point of the graph. For quadratic functions of the standard form, \(y = ax^2 + bx + c\), the vertex can be calculated using the formulas:

• The x-coordinate of the vertex, denoted as \(h\), is found with \(h = -\frac{b}{2a}\).
• The y-coordinate, also known as \(k\), is determined by substituting \(h\) back into the function, i.e., \(k = f(h)\).

In our example, the quadratic function is \(y = 2x^2 - x\). Calculating the vertex step by step, we find:

• \(a = 2\), \(b = -1\)
• \(h = \frac{-(-1)}{2*2} = \frac{1}{4}\)
• \(k = 2\left(\frac{1}{4}\right)^2 - \frac{1}{4} = \frac{1}{8} - \frac{1}{4} = -\frac{1}{8}\)
• Thus, the vertex is \((\frac{1}{4}, -\frac{1}{8})\).

Y-intercept

The y-intercept is the point where the parabola crosses the y-axis. It provides a starting point for sketching the graph of the quadratic function. This occurs when \(x = 0\). To find the y-intercept, you simply substitute \(x = 0\) into the quadratic equation. For the function \(y = 2x^2 - x\), when \(x = 0\):

• \(y = 2(0)^2 - 0 = 0\)

Hence, the y-intercept is at the origin, which is the point \((0, 0)\). It's a crucial point because it often serves as a symmetric point along with the vertex when sketching the graph.

Quadratic Function

Quadratic functions are polynomials of the second degree, generally of the form \(y = ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). The graph of a quadratic function is a parabola:

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

These functions are omnipresent in mathematics and natural phenomena due to their simple yet profound structure. Beyond just numbers, quadratic functions can easily express relationships that change over time or space, making them essential.

Graphing

Graphing a quadratic function involves plotting points to build the characteristic parabola shape. Here's how you can approach graphing step by step:

• Plot the Vertex: Start by marking the vertex \((h, k)\). For our example, it's \((\frac{1}{4}, -\frac{1}{8})\).
• Plot the Y-intercept: Next, plot the y-intercept. In our function, it's \((0, 0)\).
• Determine the Direction: Since \(a = 2\) (a positive value), the parabola opens upwards.
• Additional Points: If needed, calculate and plot additional points on either side of the vertex for accuracy.

Once these key points are plotted, draw a smooth curve through them to complete the graph of the parabola. By following these steps, the shape and position of the quadratic graph become clear and consistent.