Problem 3 - 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. $$ x=-4 y^{2}-1 $$

Step-by-Step Solution

Answer

The graph is a left-opening parabola with an x-intercept at (-1, 0) and no y-intercepts.

Step 1: Identify the Equation Type

The equation given is \( x = -4y^2 -1 \). It is a quadratic equation in terms of \( y \) and represents a parabola that opens sideways. Specifically, it opens to the left due to the negative coefficient of \( y^2 \).

Step 2: Check for Symmetries

To check for symmetry, consider substituting \( y \) with \( -y \) in the equation \( x = -4y^2 - 1 \). This substitution leaves the equation unchanged, confirming symmetry across the \( x \)-axis.

Step 3: Find the x-intercept

To find the \( x \)-intercept, set \( y = 0 \) in the equation \( x = -4y^2 - 1 \). This gives \( x = -1 \). Therefore, the \( x \)-intercept is at \((-1, 0)\).

Step 4: Find the y-intercept(s)

To find \( y \)-intercepts, set \( x = 0 \) in the equation \( x = -4y^2 - 1 \). Solving \( 0 = -4y^2 - 1 \) yields no real solutions because the left side would equal \(-1\), which cannot equal zero when squared and multiplied negatively. Therefore, there are no \( y \)-intercepts.

Step 5: Sketch the Graph

Using the information from the previous steps, plot the point \((-1, 0)\) as the \( x \)-intercept. Since the parabola opens leftwards, sketch a symmetric graph about the \( x \)-axis centered at \((-1, 0)\). It will curve leftward ensuring that it extends infinitely as \( y \) increases and decreases.

Key Concepts

Understanding Quadratic Equations

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \) when expressed in terms of \( x \). These equations typically result in a parabolic graph.
However, in our exercise, the quadratic equation is expressed differently: \( x = -4y^2 - 1 \). Here, the variable \( y \) is squared, making it a quadratic equation in \( y \), not \( x \).
This means we have a parabola, but with a focus on how \( x \) changes with \( y \). In general, the graph of a quadratic equation is a parabola, but its direction depends on the coefficient of the squared term.

• If positive, the parabola opens upward or to the right.
• If negative, like in our exercise, it opens downward or to the left.

Quadratic equations are pivotal in algebra and geometry, helping to chart curves and model numerous real-world phenomena.

Exploring Symmetry in Graphs

Symmetry is an important property of quadratic curves. In simple terms, a symmetric graph means one half is a mirror image of the other. In the context of our problem, symmetry is demonstrated along the \( x \)-axis.
To test for symmetry, we substitute \( y \) with \( -y \). For our equation, \( x = -4y^2 - 1 \), replacing \( y \) with \( -y \) produces \( x = -4(-y)^2 - 1 = -4y^2 - 1 \), which is identical to the original equation.

• This confirms the curve is symmetric about the \( x \)-axis.

Symmetry in graphs is beneficial as it allows us to understand half of the graph and infer the other half, easing the graphing process.

Finding the X-intercept

The \( x \)-intercept is the point where the graph crosses the \( x \)-axis. At this intercept, the value of \( y \) is zero. To find the \( x \)-intercept in our equation \( x = -4y^2 - 1 \), substitute \( y \) with 0.
Solving gives \( x = -1 \). Thus, the \( x \)-intercept is at \((-1, 0)\). This point is crucial as it represents where the parabola touches the \( x \)-axis.

• The intercept signifies an important boundary or starting point in graphing and analyzing the parabola.

Knowing the \( x \)-intercept helps anchor the graph, allowing you to draw it more accurately.

Absence of Y-intercept

The \( y \)-intercept is where the graph intersects the \( y \)-axis. At this point, the value of \( x \) should be zero. For the equation \( x = -4y^2 - 1 \), we attempt to find \( y \)-intercepts by setting \( x = 0 \).
This results in the equation \( 0 = -4y^2 - 1 \). However, solving this equation leads to no real solutions because \( -4y^2 \) is always negative or zero, thus \( -4y^2 - 1 \) can never equal zero.

• This implies that the graph does not cross the \( y \)-axis, therefore, it has no \( y \)-intercepts.

The absence of \( y \)-intercepts is recognized as a significant detail, guiding the understanding that the parabola is fully located on one side of the \( y \)-axis.