Problem 49 - Precalculus with Limits

Question

In Exercises 45-56, identify any intercepts and test for symmetry. Then sketch the graph of the equation. \( y = x^3 + 3 \)

Step-by-Step Solution

Answer

The x-intercept of the equation y = x^3 + 3 is -∛3, and the y-intercept is 3. The graph of the equation is neither symmetric about the y-axis nor the x-axis. With respect to the origin, it also lacks symmetry. The graph of y = x^3 + 3 is a cubic function that has been shifted up from the graph of y = x^3 by 3 units.

Step 1: Find the intercepts

Set \(y = 0\) to solve for the x-intercept: \(0 = x^3 + 3\) => \(x^3 = -3\) => \(x = -\sqrt[3]{3}\). So, the x-intercept is \(-\sqrt[3]{3}\). Set \(x = 0\) to solve for the y-intercept: \(y = 0^3 + 3 = 3\). So, the y-intercept is 3.

Step 2: Test for Symmetry

Replace \(x\) with \(-x\) in the equation \(y = x^3 + 3\), the result is \(y = -x^3 + 3\), which is not identical to the original equation. Thus, the graph is not symmetric with respect to the y-axis. Similarly, replace \(y\) with \(-y\), you get \(-y = x^3 + 3\), or \(y = -x^3 - 3\), which also differs from the original equation. Therefore, the graph does not have symmetry with respect to the x-axis either.

Step 3: Sketch the Graph

Using the x and y intercepts from Step 1 and noting that the graph has no symmetry, you should plot the intercepts and then sketch a rough curve of \(y = x^3 + 3\). Remember, the curve of \(y = x^3\) is a cubic function which increases as x increases, so the curve of \(y = x^3 + 3\) will be a shifting of this shape upward.

Key Concepts

Graph Symmetry

One way to determine if a graph is symmetric is by checking for symmetry with respect to the y-axis, x-axis, or the origin. Let's dive into each:

• Y-axis Symmetry: A graph has symmetry with respect to the y-axis if replacing each instance of \(x\) with \(-x\) in the function gives you exactly the original equation. For the cubic function \(y = x^3 + 3\), replacing \(x\) with \(-x\) results in \(y = -x^3 + 3\), which is different from the original equation. Therefore, it is not symmetric with respect to the y-axis.
• X-axis Symmetry: Similarly, a graph has x-axis symmetry if by replacing \(y\) with \(-y\) gives you the original equation. Doing this to \(y = x^3 + 3\) results in \(-y = x^3 + 3\), which solved for y is \(y = -x^3 - 3\). This does not match the original function, hence no x-axis symmetry.
• Origin Symmetry: To check for symmetry about the origin, replace \(x\) with \(-x\) and \(y\) with \(-y\). This transformation yields \(-y = -x^3 + 3\) for this specific function. Solving for \(y\) results in \(y = x^3 - 3\), which fails to match the original equation as well. Thus, there's no origin symmetry either.

These steps confirm that the graph of \(y = x^3 + 3\) possesses no symmetry.

Intercepts

Intercepts are points where the graph cuts through the axes, providing valuable anchors for sketching graphs. Let's find them:

• X-intercept: Set \(y = 0\) in the equation to find where the curve intersects the x-axis. Doing so gives us the equation \(0 = x^3 + 3\), simplifying to \(x^3 = -3\), hence \(x = -\sqrt[3]{3}\). This means the graph crosses the x-axis at \((-\sqrt[3]{3}, 0)\).
• Y-intercept: Here, set \(x = 0\) in the original equation to find where the curve intersects the y-axis. The calculation is simple: \(y = 0^3 + 3 = 3\). Accordingly, the y-intercept is at \((0, 3)\).

The intercepts offer direct insight into where the graph intersects the axes, making the understanding and sketching of the graph more intuitive.

Graph Sketching

After identifying intercepts and examining symmetry, sketching the graph becomes much more straightforward:

• Begin by plotting the intercepts: \(-\sqrt[3]{3}\) on the x-axis and \(3\) on the y-axis serve as tangible starting points.
• Recall the basic shape of the cubic function \(y = x^3\), which naturally curves and exhibits steep ascend and descend as \(x\) moves away from zero. The classic cubic curve asymmetrically rises from negative to positive infinity.
• Add the shift of \(+3\). This adjustment indicates the entire graph is moved upward by three units, so the central curve of this cubic function now behaves as \(y = x^3 + 3\).

With intercepts plotted and curve shape understood, sketching the graph becomes a matter of integrating these elements into a cohesive picture.