Problem 2 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The graph is a horizontal parabola, opening left, with vertex at (1,0) and symmetric about the x-axis.

Step 1: Determine the Type of Equation

The equation given is \(x = -y^2 + 1\). This equation describes a parabola, but note that it is given in the form of \(x\) as a function of \(y\), which means it opens horizontally rather than vertically.

Step 2: Identify Symmetries

To identify symmetries, we analyze the equation for even or odd terms. On changing \(y\) to \(-y\), the equation becomes \(x = -(-y)^2 + 1 = -y^2 + 1\), which is the same as the original, indicating symmetry about the x-axis.

Step 3: Find the y-intercepts

y-intercepts occur when \(x = 0\). Set \(-y^2 + 1 = 0 \) and solve for \(y\):\[-y^2 = -1 \implies y^2 = 1 \implies y = \pm 1\]. Thus, the y-intercepts are at \((0, 1)\) and \((0, -1)\).

Step 4: Find the x-intercept

x-intercepts occur when \(y = 0\). Plug \(y = 0\) into the equation: \(x = -(0)^2 + 1 = 1\). Thus, the x-intercept is at \((1, 0)\).

Step 5: Graph the Equation

When plotting \(x = -y^2 + 1\), note that it is a horizontal parabola that opens towards the left (since the coefficient of \(y^2\) is negative). It is symmetric about the x-axis, passing through the points \((1, 0)\), \((0, 1)\), and \((0, -1)\). Draw the parabola with the vertex at \((1, 0)\) and extending through these points.

Key Concepts

Symmetry in Equations

Symmetry in equations helps us understand the shape and orientation of a graph. It tells us if one side of the graph mirrors the other, making it easier to visualize the equation without plotting numerous points. Consider the equation \(x = -y^2 + 1\). To check for symmetry, we substitute \(y\) with \(-y\). This substitution doesn't change the equation as it remains \(x = -y^2 + 1\), demonstrating symmetry about the x-axis. When an equation is symmetric about the x-axis, any point \((x, y)\) on the graph also has a mirrored point at \((x, -y)\). This implies that the graph can be folded along the x-axis, and both halves will match perfectly. Symmetry is a property that simplifies graph making because it reduces the number and difficulty of calculations required to verify graph points.

Parabolic Equations

Parabolic equations form curves known as parabolas. Typically, these equations are expressed in the form \(y = ax^2 + bx + c\). However, when dealing with \(x = -y^2 + 1\), the equation is a parabola but in a horizontal format, meaning it opens sideways.Key points about horizontal parabolas:

• They open to the left if the coefficient of \(y^2\) is negative.
• They open to the right if the coefficient of \(y^2\) is positive.

For \(x = -y^2 + 1\), the negative coefficient of \(y^2\) indicates that the parabola opens to the left. Such parabolas usually have a vertex, the highest or lowest point on the curve, at the point where the parabola begins to bend. Here, the vertex is at the point \((1, 0)\), serving as the starting point from which the parabola stretches out symmetrically along the x-axis.

X and Y Intercepts

Intercepts are points where the graph crosses the axes, providing us with valuable insights into the behavior of an equation.**Y-intercepts** occur where the graph crosses the y-axis. For the equation \(x = -y^2 + 1\), we find the y-intercepts by setting \(x = 0\) and solving for \(y\):\[-y^2 + 1 = 0 \implies y^2 = 1 \implies y = \pm 1\].This gives us y-intercepts at \((0, 1)\) and \((0, -1)\).**X-intercepts** occur where the graph crosses the x-axis. Set \(y = 0\) in the equation:\[x = -(0)^2 + 1 = 1\].Therefore, the x-intercept is at \((1, 0)\).Knowing intercepts can support the graphing process, as these points act as anchors or verifications when sketching the overall shape of the equation's graph. They highlight essential starting or stopping coordinates in the movement of the curves across the axis.