Problem 5

Question

sketch the graph of each function. Do not use a graphing calculator. (Assume the largest possible domain.) $$ y=-2 x^{2}-3 $$

Step-by-Step Solution

Verified

Answer

The graph is a downward-opening parabola with vertex (0, -3). Plot additional points for accuracy.

1Step 1: Identify the Function

The given function is a quadratic function in the form $ y = -2x^2 - 3 $, where the coefficient of $ x^2 $ is negative, indicating it is a downward-opening parabola.

2Step 2: Determine the Vertex

For the quadratic $ y = ax^2 + bx + c $, the vertex's x-coordinate can be found using $ x = -\frac{b}{2a} $. In this function, $ a = -2 $, $ b = 0 $, and $ c = -3 $, giving $ x = 0 $. Substituting $ x = 0 $ into the function gives $ y = -3 $. Thus, the vertex is at $ (0, -3) $.

3Step 3: Find the Axis of Symmetry

The axis of symmetry for a parabola in the form $ y = ax^2 + bx + c $ is the vertical line $ x = -\frac{b}{2a} $. For this function, it is $ x = 0 $, which corresponds to the x-coordinate of the vertex.

4Step 4: Find the Y-intercept

The y-intercept is found by setting $ x = 0 $. Substituting gives $ y = -3 $, so the y-intercept is $ (0, -3) $, which coincides with the vertex.

5Step 5: Calculate Additional Points

To better sketch the graph, find additional points. Let's choose $ x = 1 $ and $ x = -1 $. For $ x = 1 $, $ y = -2(1)^2 - 3 = -5 $. For $ x = -1 $, $ y = -2(-1)^2 - 3 = -5 $. Therefore, we have points $ (1, -5) $ and $ (-1, -5) $.

6Step 6: Sketch the Parabola

Plot the vertex $ (0, -3) $, y-intercept $ (0, -3) $, and additional points $ (1, -5) $ and $ (-1, -5) $. Draw a smooth curve through these points, making sure the parabola opens downward due to the negative sign of the coefficient $ -2 $.

Key Concepts

Vertex of a ParabolaAxis of SymmetryY-intercept

Vertex of a Parabola

The vertex of a parabola is a critical point that provides key information about the graph of a quadratic function. In any quadratic function given by the general form $ y = ax^2 + bx + c $, the vertex is the point where the parabola reaches its maximum or minimum value. This depends on whether the parabola opens upwards or downwards.

For a parabola that opens downwards, the vertex is the highest point, called the maximum point.
For a parabola that opens upwards, the vertex is the lowest point, known as the minimum point.

To find the vertex, we use the formula for the x-coordinate: $ x = -\frac{b}{2a} $. Once we determine the x-coordinate, we substitute it back into the quadratic function to find the y-coordinate. In the exercise provided, the quadratic function is $ y = -2x^2 - 3 $. With $ a = -2 $, $ b = 0 $, and $ c = -3 $, the x-coordinate of the vertex is calculated as $ x = -\frac{0}{2(-2)} = 0 $. Subsequently, we find the y-coordinate by substituting $ x = 0 $ into the equation, giving $ y = -3 $. Thus, the vertex is at the point $(0, -3)$, which is also the highest point on the graph since the parabola opens downward.

Axis of Symmetry

The axis of symmetry of a parabola is a vertical line that divides the parabola into two mirror-image halves. This line passes through the vertex of the parabola, ensuring that one side is a reflected version of the other. The symmetrical nature of the parabola means that for any point $(x, y)$ on one side of the axis of symmetry, there is a corresponding point $(-x, y)$ on the opposite side. For any quadratic function in the form $ y = ax^2 + bx + c $, the equation for the axis of symmetry is given by:$ x = -\frac{b}{2a} $. This formula is derived from the coordinate of the vertex and is identical to the calculation for the x-coordinate of the vertex. In our case with the function $ y = -2x^2 - 3 $, the axis of symmetry coincides with $ x = 0 $, aligning with the vertex's x-coordinate. This means that the line $ x = 0 $ is the axis of symmetry, which means the left side and right side of the parabola on either side of $ x = 0 $ are mirror images of each other.

Y-intercept

A y-intercept is a specific point on the graph where the curve crosses the y-axis. At the y-intercept, the x-coordinate is always zero. Finding the y-intercept of a quadratic function is straightforward: simply substitute $ x = 0 $ into the equation and solve for $ y $. In the provided exercise with the function $ y = -2x^2 - 3 $, plugging $ x = 0 $ into the equation yields $ y = -3 $. Therefore, the y-intercept is $(0, -3)$. It is noteworthy that in a parabola, the vertex and y-intercept can sometimes be the same point, particularly when the axis of symmetry is $ x = 0 $, which centers the parabola over the y-axis, as seen in the exercise.

The y-intercept gives us a concrete point through which we start plotting the parabola.
It helps in visualizing the graph since it's part of the most basic framework used to sketch the parabola.

The y-intercept $(0, -3)$ is a vital starting point to understand how the parabola behaves on the graph.

Problem 4

Problem 5

Other exercises in this chapter

Problem 4

State the range for the given functions. Graph each function. $$ f(x)=x^{2},-\frac{1}{2}

View solution

Problem 4

Solve the following equations: (a) $|2 x+4|=|5 x-2|$ (b) $|1+2 u|=|5-u|$ (c) $\left|4+\frac{1}{2}\right|=\left|\frac{3}{2} t-2\right|$ (d) \(|2 s-6|=|3-s|

View solution

Problem 5

(a) Show that, for $x \neq 1$, $$ \frac{x^{2}-1}{x+1}=x-1 $$ (b) Are the functions $f(x)$ and $g(x)$ equal? $$ \begin{array}{l} f(x)=\frac{x^{2}-1}{x+1},

View solution

Problem 5

Solve the following inequalities: (a) $|5 x-2| \leq 4$ (b) $|3-4 x|>8$ (c) $|7 x+4| \geq 3$ (d) \(|3+2 x|

View solution