Problem 16 - Basic Technical Mathematics with Calculus

Question

Sketch the graph of each parabola by using the vertex, the \(y\) -intercept, and two other points, not including the \(x\) -intercepts. Check the graph using a calculator. \(y=-3 x^{2}-x\)

Step-by-Step Solution

Answer

The vertex is \(\left(\frac{1}{6}, -\frac{1}{4}\right)\), y-intercept is \((0, 0)\), other points \((1, -4)\), \((-1, -2)\); graph is a downward parabola.

Step 1: Identify the Vertex of the Parabola

The general form of a quadratic equation is \(y = ax^2 + bx + c\). In this case, \(a = -3\), \(b = -1\), and \(c = 0\). The vertex \((h, k)\) for a parabola in the form \(ax^2 + bx + c\) is given by \(h = -\frac{b}{2a}\) and \(k = f(h)\). Calculate \(h\): \(h = -\frac{-1}{2 \times (-3)} = \frac{1}{6}\). Now, substitute \(x = \frac{1}{6}\) into the equation to find \(k\): \(k = -3\left(\frac{1}{6}\right)^2 - \frac{1}{6} = -\frac{1}{12} - \frac{1}{6} = -\frac{3}{12} = -\frac{1}{4}\). Thus, the vertex is \(\left(\frac{1}{6}, -\frac{1}{4}\right)\).

Step 2: Find the y-intercept of the Parabola

The y-intercept occurs when \(x = 0\). Substitute \(x = 0\) into the equation \(y = -3x^2 - x\) to find \(y\). \(y = -3(0)^2 - 0 = 0\). Therefore, the y-intercept is \((0, 0)\).

Step 3: Choose Two Additional Points

We need to find two other points by selecting two different \(x\)-values. Let's choose \(x = 1\) and \(x = -1\). For \(x = 1\): substitute into the equation to get \(y = -3(1)^2 - 1 = -3 - 1 = -4\), so the point is \((1, -4)\). For \(x = -1\): substitute into the equation to get \(y = -3(-1)^2 - (-1) = -3 + 1 = -2\), so the point is \((-1, -2)\).

Step 4: Sketch the Graph of the Parabola

Now use all points obtained to sketch the graph. Start by marking the vertex \(\left(\frac{1}{6}, -\frac{1}{4}\right)\), then plot the y-intercept \((0, 0)\), and the additional points \((1, -4)\) and \((-1, -2)\). Since the parabola opens downwards (because \(a = -3\), which is negative), sketch the curve passing through these points forming a "U"-shape.

Step 5: Verify the Graph with a Calculator

To ensure accuracy, plot the equation \(y = -3x^2 - x\) using a graphing calculator. The graph should match your sketch, showing a downward-opening parabola with the plotted vertex, intercept, and additional points. Adjust your graph as needed to align perfectly with the calculator.

Key Concepts

Vertex Form of a Parabola

A parabola is a symmetric curve on a graph described by a quadratic equation of the form \( y = ax^2 + bx + c \). To sketch a parabola accurately, it's crucial to find its vertex, which is the highest or lowest point on the graph depending on the orientation. The vertex can be found using the formula:

• \( h = -\frac{b}{2a} \)
• \( k = f(h) \)

Here, \( h \) is the x-coordinate and \( k \) is the y-coordinate of the vertex. By substituting \( h \) into the original equation, you'll find \( k \). For the equation \( y = -3x^2 - x \), we calculate \( h \) as \( \frac{1}{6} \) and \( k \) as \( -\frac{1}{4} \), resulting in the vertex \( \left(\frac{1}{6}, -\frac{1}{4}\right) \). This point is useful for graphing as it helps to define the shape and position of the parabola on the graph.

Y-Intercept

The y-intercept of a parabola is the point where the graph crosses the y-axis. It is found by solving the quadratic equation when \( x = 0 \). This is because any point on the y-axis has an x-coordinate of zero. For our specific equation \( y = -3x^2 - x \), substituting \( x = 0 \) gives:
\( y = -3(0)^2 - 0 = 0 \).
This means the y-intercept is at the origin, or the point \( (0, 0) \). The y-intercept is another key point used when sketching the graph because it provides a starting point for drawing the parabola's path.

Graphing Parabolas

Graphing a parabola involves plotting its key points, such as the vertex and the y-intercept, and then drawing the curve through these points. Let's break down the steps:

• Plot the vertex \( \left(\frac{1}{6}, -\frac{1}{4} \right) \).
• Mark the y-intercept \( (0, 0) \).

Next, determine additional points by choosing random \( x \)-values and computing their corresponding \( y \)-values. This is where plot symmetry is handy:

• For \( x = 1 \), \( y = -4 \), point: \( (1, -4) \).
• For \( x = -1 \), \( y = -2 \), point: \( (-1, -2) \).

Once these points are plotted, they help form the parabolic "U"-shape. Remember, because \( a = -3 \) is negative, the parabola opens downwards.

Additional Points on Parabola

To ensure you have enough information to accurately sketch a parabola, it's helpful to choose additional points beyond the vertex and y-intercept. This allows you to capture the curve's full shape.
In our approach:

• We selected two additional x-values: \( x = 1 \) and \( x = -1 \).
• For \( x = 1 \), substituting into \( y = -3(1)^2 - 1 \) gives \( y = -4 \), resulting in the point \( (1, -4) \).
• For \( x = -1 \), substituting gives \( y = -2 \), resulting in the point \( (-1, -2) \).

Adding these additional points makes the graph more accurate and visually precise, ensuring the "U" shape of the downward-opening parabola is clearly represented. This coverage of points creates a clearer path for sketching and verifying via a graphing tool.