Problem 107 - A Graphical Approach to Precalculus with Limits

Question

Graph by hand the equation of the circle or the parabola with a horizontal axis. $$x=-(y+1)^{2}+2$$

Step-by-Step Solution

Answer

It's a left-opening parabola with vertex at (2, -1).

Step 1: Identify the Type of Equation

The given equation is \( x = -(y+1)^2 + 2 \). This is an equation of a parabola and not a circle because it can be expressed in the form \( x = a(y-k)^2 + h \).

Step 2: Identify the Direction of Opening

Since the coefficient of \((y+1)^2\) is negative, the parabola opens to the left. Parabolas in the form \( x = -(y+1)^2 + 2 \) open leftward when the quadratic term is \( (y+1)^2 \).

Step 3: Identify the Vertex

The standard form of a horizontal parabola is \( x = a(y-k)^2 + h \). For the equation \( x = -(y+1)^2 + 2 \), comparing it to the standard form gives \( h = 2 \) and \( k = -1 \). Therefore, the vertex of the parabola is at the point \( (2, -1) \).

Step 4: Determine Key Points

To draw the parabola accurately, choose some y-values near the vertex, say \( y = -3, -2, 0, 1 \), and substitute these into the equation to find corresponding x-values:\( y = -3 \Rightarrow x = -(3-1)^2 + 2 = -4 + 2 = -2 \); \( y = -2 \Rightarrow x = -(-2 + 1)^2 + 2 = -1 + 2 = 1 \); \( y = 0 \Rightarrow x = -(0+1)^2 + 2 = -1 + 2 = 1 \); \( y = 1 \Rightarrow x = -(1+1)^2 + 2 = -4 + 2 = -2 \).

Step 5: Plot the Points and Draw the Parabola

On a coordinate plane, plot the vertex at \((2, -1)\) and the other points you calculated: \((-2, -3), (1, -2), (1, 0), (-2, 1)\). Connect these points smoothly to form the parabola, ensuring that it opens to the left.

Key Concepts

Equation of a Parabola

When graphing parabolas, you first need to understand the equation you're dealing with. A standard equation for a parabola can either follow a vertical form, \( y = a(x-h)^2 + k \), or a horizontal one, \( x = a(y-k)^2 + h \). In our problem, we have the horizontal form, specifically \( x = -(y+1)^2 + 2 \). This appearance as \( x = a(y-k)^2 + h \) signifies that it indeed describes a parabola.

Key components in this form include:

• \( a \): Determines the width and direction of the parabola. If \( a \) is negative, the parabola opens in the opposite direction (left for horizontal).
• \( (h, k) \): The vertex of the parabola. This is the "turning point" of the curve.

Understanding the equation's structure helps in correctly graphing and analyzing the shape.

Vertex of a Parabola

The vertex of a parabola is a crucial point that serves as a central feature of its graph. For any horizontal form \( x = a(y-k)^2 + h \), the vertex is located at \( (h, k) \).

In the given equation, \( x = -(y+1)^2 + 2 \), the vertex is found by identifying \( h \) and \( k \) from the equation. Here:

• \( h = 2 \)
• \( k = -1 \)

Thus, the vertex is at \( (2, -1) \).

This point essentially acts as the "pivot" of your parabola, marking the maximum or minimum height (or width, in horizontal cases) reached by the curve. Plotting the vertex correctly is a critical step in sketching the parabola.

Direction of Parabola Opening

To determine the direction of a parabola's opening, examine the coefficient \( a \) of the squared term in its equation. For the standard horizontal form \( x = a(y-k)^2 + h \), if \( a \) is positive, the parabola opens to the right. Conversely, if \( a \) is negative, as in our equation \( x = -(y+1)^2 + 2 \), it opens to the left.

The negative sign in front of the squared term \((y+1)^2\) indicates that our parabola does not extend rightward but instead curves left.

Grasping which way your parabola curls is significant for both plotting and understanding the implications of real-world problems represented by such mathematical models. Always keep an eye on the sign of \( a \) to avoid mistakes during graphing.