Problem 5 - 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. $$ x^{2}+y=0 $$

Step-by-Step Solution

Answer

The graph is a downward-opening parabola, \(y = -x^2\), symmetric about the y-axis with vertex at the origin \((0,0)\).

Step 1: Check for Symmetry

To check if the equation \(x^2 + y = 0\) has symmetry about the x-axis, y-axis, and origin, we substitute \(-x\), \(-y\), and \(-x, -y\) into the equation and see if the equation remains the same. - Substitute \(-x\): \((-x)^2 + y = x^2 + y = 0\). The equation remains the same.- Substitute \(-y\): \(x^2 - y = 0\) which is not the same as the original equation.- Substitute \(-x, -y\): \((-x)^2 - y = x^2 - y = 0\) which is not the same as the original equation.Thus, the graph is symmetric about the y-axis.

Step 2: Identify the x-Intercepts

To find the x-intercepts, set \(y = 0\) in the equation and solve for \(x\):\[x^2 + 0 = 0\]\[x^2 = 0\]\[x = 0\]The x-intercept of the graph is \((0, 0)\).

Step 3: Identify the y-Intercepts

To find the y-intercepts, set \(x = 0\) in the equation and solve for \(y\):\[0^2 + y = 0\]\[y = 0\]The y-intercept of the graph is \((0, 0)\).

Step 4: Plot and Sketch the Graph

Given that we have found the graph to be symmetric about the y-axis and the intercept point at \((0,0)\), we can plot the graph. Notice that the equation \(x^2 + y = 0\) or equivalently \(y = -x^2\) represents a parabola opening downwards. The vertex of the parabola is at the origin \((0,0)\). Simply sketch the parabola with the vertex at the origin and opening downwards.

Key Concepts

Symmetry in Graphs

Understanding symmetry in graphs helps simplify the process of graphing equations. A graph is symmetric if it looks the same after a specific transformation: around the x-axis, the y-axis, or the origin. For our equation, let's examine how transformations affect it.

• To test for symmetry about the x-axis, replace y with -y and see if the equation remains unchanged.
• To test for symmetry about the y-axis, replace x with -x and check if the equation stays the same.
• For symmetry about the origin, replace x with -x and y with -y, then observe the equation's consistency.

In this case, the equation is symmetric about the y-axis because replacing x with -x does not change the equation. Symmetry about the y-axis means if the graph is folded over the y-axis, both sides will overlap, making graphing easier as you only need to plot one side and replicate the reflected points on the other.

X-Intercepts

X-intercepts are vital points where the graph crosses the x-axis, making y equal to zero at these points. To find the x-intercepts of our equation, set y to zero and solve for x. Here’s how you do it:

• Substitute y = 0 into the equation: \(x^2 + 0 = 0\) simplifies to \(x^2 = 0\).
• Solving for x gives \(x = 0\).

Thus, the only x-intercept is at the origin, \((0, 0)\). Since there is only one point, it also confirms our graph is centered on this point.

Y-Intercepts

Just like x-intercepts, y-intercepts are where the graph meets the y-axis, which occurs when x equals zero. To find the y-intercepts, let's put x = 0 in the equation and solve for y.

• Substitute x = 0: \(0^2 + y = 0\) simplifies to \(y = 0\).

The y-intercept is again the point \((0, 0)\), aligning with the x-intercept. This overlap of intercepts confirms the graph passes through the origin, a key feature for sketching the graph of the equation.

Parabolas

A parabola is a curve described by quadratic equations like \(ax^2 + bx + c\). In our case, the equation \(x^2 + y = 0\) can be rearranged to \(y = -x^2\), showcasing its parabolic nature. Parabolas can open upwards or downwards depending on the coefficient of the \(x^2\) term.

• Here, the equation \(y = -x^2\) denotes a parabola that opens downwards because the coefficient of \(x^2\) is negative.
• The vertex of this parabola is the peak point, located at \((0,0)\).

Knowing that the vertex is at the origin and the direction of opening, you can accurately draw the curve. Understanding the properties of parabolas helps in visualizing the symmetry and intercepts discussed earlier.