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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.

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 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.

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.