Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-30, plot the graph of each equation. Begin by checking for symmetries and be sure to find all \(x\) - and \(y\)-intercepts. $$ 2 x^{2}-4 x+3 y^{2}+12 y=-2 $$

Step-by-Step Solution

Answer

Symmetric about y-axis; x-intercept is (1,0), y-intercepts are approximately (0,0.91) and (0,-4.91).

Step 1: Rearrange the Equation

We start with the given equation and rearrange terms: \(2x^2 - 4x + 3y^2 + 12y = -2\). Move all terms to one side so the equation becomes \(2x^2 - 4x + 3y^2 + 12y + 2 = 0\).

Step 2: Check for Symmetries

To find symmetries, we check if replacing \(x\) with \(-x\) results in the same equation. Doing so gives no change in the equation, thus it's symmetric with respect to the \(y\)-axis. Similarly, replacing \(y\) with \(-y\) does not preserve the equation, thus it is not symmetric with respect to the \(x\)-axis. Checking both \(x\) and \(y\) at once also results in a change, thus it is not symmetric about the origin.

Step 3: Find the x-intercepts

To find where the graph crosses the x-axis, we set \(y = 0\) in the equation. Simplifying, we have \(2x^2 - 4x = -2\) or \(2x^2 - 4x + 2 = 0\). Divide the equation by 2: \(x^2 - 2x + 1 = 0\), which factors as \((x-1)^2 = 0\). So the x-intercept is \((1, 0)\).

Step 4: Find the y-intercepts

To find the y-intercepts, set \(x = 0\). The equation becomes \(3y^2 + 12y + 2 = 0\). Use the quadratic formula for \(y\): \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a=3\), \(b=12\), \(c=2\). This results in \(y = \frac{-12 \pm \sqrt{144 - 24}}{6} = \frac{-12 \pm \sqrt{120}}{6}\). Simplified, the solutions are \(y = -2 + \frac{2\sqrt{30}}{3}\) and \(y = -2 - \frac{2\sqrt{30}}{3}\). Thus, y-intercepts are approximately \((0, 0.91)\) and \((0, -4.91)\).

Step 5: Verify Other Critical Points

Further verify intersections and key points by rearranging the equation in terms of \(x\) or \(y\), and re-calculate any points of interest. Check against the intercepts and symmetry found.

Key Concepts

Symmetry in Graphs

Graph symmetry is an important concept that makes understanding the behavior of graphs much easier. A graph is said to have symmetry if a transformation applied to it results in a graph that appears unchanged. For the equation \(2x^2 - 4x + 3y^2 + 12y = -2\), we need to consider three types of symmetries:

• y-axis symmetry: To check for this, replace \(x\) with \(-x\). If the equation remains unchanged, then it is symmetric about the y-axis. For our equation, replacing \(x\) with \(-x\) yields the same equation, confirming it is y-axis symmetric.
• x-axis symmetry: Swap \(y\) with \(-y\). If the equation is unchanged, the graph is symmetric about the x-axis. In this case, replacing \(y\) with \(-y\) changes the equation, so it has no x-axis symmetry.
• Origin symmetry: Replace both \(x\) and \(y\) with \(-x\) and \(-y\), respectively. If unchanged, the graph has origin symmetry. Our equation alters, indicating no symmetry about the origin.

Recognizing symmetry can simplify graphing processes and aid in predicting graph behaviors at different points.

x-intercepts

The x-intercepts of a graph are points where the graph crosses the x-axis. At these points, the value of \(y\) is zero. To find them, set \(y = 0\) in the equation and solve for \(x\). For our example, setting \(y = 0\) simplifies the equation to \(2x^2 - 4x = -2\) or \(2x^2 - 4x + 2 = 0\). Divide through by 2 to get \(x^2 - 2x + 1 = 0\), which can be rewritten in factored form as \((x - 1)^2 = 0\). This reveals a double root at \(x = 1\), indicating that there is one unique x-intercept point where the graph just touches or rebounds from the x-axis, namely at \((1,0)\). Double roots suggest that the graph is tangent to the axis at this intercept. Understanding whether an intercept is a simple or double root can help in sketching the graph and understanding its structure at x-intercept positions.

y-intercepts

Finding y-intercepts involves determining where the graph crosses the y-axis. At these points, the value of \(x\) is zero. So, substitute \(x = 0\) into the equation and solve for \(y\). In our equation \(2x^2 - 4x + 3y^2 + 12y + 2 = 0\), setting \(x = 0\) gives \(3y^2 + 12y + 2 = 0\).Using the quadratic formula, \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=3\), \(b=12\), and \(c=2\), we find:

• Calculate discriminant: \(b^2 - 4ac = 144 - 24 = 120\)
• Two solutions: \(y = \frac{-12 \pm \sqrt{120}}{6}\)
• Simplify to get two decimal solutions as \(y \approx 0.91\) and \(y \approx -4.91\)

Thus, the y-intercepts are approximately \((0, 0.91)\) and \((0, -4.91)\). These values show where the graph crosses the y-axis, providing essential points for graphing.

Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It allows us to find the x-values where these curves touch or cross the axes. The formula is given as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Understanding how and when to use it is key:

• Roots: The solutions to the quadratic equation are given by the formula, representing points where the curve crosses the axis.
• Discriminant: The term under the square root, \(b^2 - 4ac\), is the discriminant which indicates the nature of the roots - real and distinct, real and equal (double root), or complex.
• Simplicity: This formula provides a reliable method to find intercepts, allowing us to graph sections of the curve accurately.

Overall, the quadratic formula helps not only in finding intercepts accurately but also in verifying other critical points when graphing or analyzing quadratic curves.