Question

Graph the quadratic equation. Label the vertex and axis of symmetry.\(y=-3 x^2-2\)

Short answer

The vertex is at (0,-2) and the axis of symmetry is at \(x=0\). When these are plotted on the graph along with a few other points determined by the equation \(y = -3x^2 - 2\), a downward-opening parabola is formed.

Step-by-step solution

Identify the Vertex

The standard form of a quadratic equation is \(y = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola. For equation \(y = -3x^2 - 2\), the equation is already in the form \(y = ax^2 + k\), which means the vertex \((h, k)\) is \((0, -2)\).

Find the Axis of Symmetry

The axis of symmetry of a parabola represented by \(y = a(x-h)^2 + k\) is \(x=h\). Since the vertex of this equation is (0,-2), the axis of symmetry would be \(x=0\).

Plot the Vertex and Axis of Symmetry

Start by plotting the vertex (0,-2). Then, draw a vertical line through the vertex to represent the axis of symmetry, which is \(x=0\).

Plot Other Points and Draw the Parabola

Now choose a few more points on either side of the axis of symmetry and find their corresponding y-values using the equation \(y = -3x^2 - 2\). This will give you additional points to help complete the parabolic shape. Lastly, connect all points to graph the parabola, making sure the shape is symmetric around the axis.

Key concepts

Vertex of a Parabola

Understanding the vertex of a parabola is essential when graphing quadratic equations. The vertex represents the highest or lowest point on the parabola, depending on whether the parabola opens upwards or downwards.

In the given equation, the form does not directly reveal the vertex as it is not in the vertex form of a quadratic, which is \(y = a(x-h)^2 + k\). However, since the equation is \(y = -3x^2 - 2\), we can tell that the parabola opens downward (because the coefficient of \(x^2\) is negative) and the vertex is at the origin of the coordinate system, which is the point \(0, -2\).

The vertex's x-coordinate is the same as the axis of symmetry, which in this case is \(x=0\), and the y-coordinate is the value of the equation when we substitute \(x\) with the x-coordinate of the vertex. In our example, substituting \(x\) with 0 gives us the y-coordinate of the vertex: \(y = -3(0)^2 - 2 = -2\). Hence, the vertex is \(\left(0, -2\right)\).

When graphing, we start by marking the vertex on the graph, ensuring that the rest of the parabola is shaped correctly about this point.

Axis of Symmetry

The axis of symmetry in a parabola is a vertical line that bisects the parabola into mirror images on either side. This axis runs through the vertex and every parabola has one axis of symmetry. In the standard form \(y = a(x-h)^2 + k\), the line of symmetry is described by the equation \(x = h\).

For the quadratic equation \(y = -3x^2 - 2\), the axis of symmetry corresponds to the x-coordinate of the vertex since the equation is equivalent to \(y = -3(x-0)^2 - 2\). Thus, the axis of symmetry is \(x=0\), which is simply the y-axis. Graphically, this means you will draw a straight line along the y-axis to indicate the axis of symmetry, ensuring that the parabola on either side is a reflection of the other.

When graphing more points, make sure to select points symmetrically located relative to the axis of symmetry to confirm that the curvature of your parabola is accurate.

Standard Form of a Quadratic Equation

The standard form of a quadratic equation is expressed as \(y = ax^2 + bx + c\). This form is extremely useful in identifying the direction in which the parabola opens and its y-intercept directly.

For the given equation, \(y = -3x^2 - 2\), it's in standard form with \(a = -3\), \(b = 0\), and \(c = -2\). The coefficient \(a\) indicates that the parabola opens downward because it is negative. The coefficient \(b\) is the term that determines the shape of the parabola, and since it's 0, there's no linear component affecting the symmetry of the graph. Lastly, the constant term \(c\) is the y-intercept, which tells us the point where the parabola crosses the y-axis. In this case, the y-intercept and the vertex both lie at \(0, -2\) as there is no horizontal shift.

When transforming this standard form to a more graph-friendly version, such as vertex form, one might use completing the square or other methods. However, in this scenario, the vertex is immediately identifiable without such transformation, simplifying the graphing process.