Problem 11 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-20: Sketch the graph of the equation, and label the \(x\) - and \(y\)-intercepts. $$ x=-y^{2}+3 $$

Step-by-Step Solution

Answer

The graph is a sideways parabola with a vertex at (3, 0) and intercepts at (3, 0), (0, \(\sqrt{3}\)), and (0, -\(\sqrt{3}\)).

Step 1: Understand the Equation

The equation given is \( x = -y^2 + 3 \). This is a quadratic equation in \( y \). In standard form, it looks like \( x = -y^2 + 3 \). Let's analyze it to identify the type of function and its characteristics.

Step 2: Determine the Shape and Direction

The equation resembles a parabola with the 'x' in terms of 'y'. Since the coefficient of \( y^2 \) is negative, the parabola opens to the left. In this case, it is horizontal because \( x \) is expressed in terms of \( y^2 \).

Step 3: Find the Vertex of the Parabola

The equation can be rewritten as \( x = 3 - y^2 \), which indicates the vertex of the parabola is at \( (3, 0) \) because it is already in vertex form for a sideways parabola. Here, \( h = 3 \) and \( k = 0 \). The vertex is at the maximum point in terms of \( x \).

Step 4: Calculate the Intercepts

For the \( x \)-intercept, set \( y = 0 \) and solve for \( x \): \( x = -0^2 + 3 = 3 \). Thus, the \( x \)-intercept is \( (3, 0) \). For the \( y \)-intercepts, set \( x = 0 \) and solve for \( y \):\[0 = -y^2 + 3 \y^2 = 3 \y = \pm \sqrt{3}\]So there are two \( y \)-intercepts: \( (0, \sqrt{3}) \) and \( (0, -\sqrt{3}) \).

Step 5: Sketch the Graph

Plot the vertex at \( (3, 0) \) and the intercepts \( (3, 0) \), \( (0, \sqrt{3}) \), \( (0, -\sqrt{3}) \). Sketch a symmetric parabola opening to the left through these points. Make sure the graph reflects the sideways nature of the parabola.

Key Concepts

Equation of a Parabola

The equation of a parabola is a crucial concept in algebra and geometry, representing the path that a quadratic function creates. Typically, it's written in standard form as either

• \( y = ax^2 + bx + c \) for a vertical parabola
• \( x = ay^2 + by + c \) for a horizontal parabola.

The given exercise provides us with the equation \( x = -y^2 + 3 \). Here, the variable \( x \) is expressed in terms of \( y^2 \), indicating we are dealing with a "sideways" parabola. The general form for a sideways parabola is \( x = a(y-k)^2 + h \), where \((h, k)\) represents the vertex of the parabola. The key characteristics of this equation are determined by the coefficients. In our equation, the absence of a linear \( y \) term shows the parabola does not tilt, and the negative sign indicates that it opens to the left.

x-intercept

The x-intercept of a graph is the point where the graph crosses the x-axis. For these intercepts, we set the other variable (in this problem, \( y \)) to zero and solve for \( x \). For this exercise, with the equation \( x = -y^2 + 3 \), simple substitution provides the answer:

• Setting \( y = 0 \) results in \( x = 3 \).

Thus, the x-intercept is at (3, 0). This is a critical point to plot when sketching the graph, showing precisely where the parabola touches or crosses the x-axis.

y-intercept

Similarly, the y-intercept is where the graph intersects the y-axis. This point is found by setting \( x = 0 \) and solving for \( y \) in the given equation. Utilizing the equation \( x = -y^2 + 3 \):

• Set \( x = 0 \): \( 0 = -y^2 + 3 \)
• This simplifies to \( y^2 = 3 \), resulting in \( y = \pm \sqrt{3} \).

Thus, we identify two y-intercepts: one at (0, \( \sqrt{3} \)) and the other at (0, \( -\sqrt{3} \)). These points indicate where the parabola intersects the y-axis and are necessary for an accurate graph.

Vertex of a Parabola

The vertex of a parabola serves as its "tip" or maximum/minimum point, depending on the parabola's orientation and direction. For a sideways parabola like ours, \( x = -y^2 + 3 \), we find the vertex where the square term \((y^2)\) equals zero, maintaining the maximum \( x \) value.

• From the equation \( x = 3 - y^2 \), we determine the vertex at \((3, 0)\).

This vertex acts as a central point for both plotting and understanding the parabola's symmetrical structure. In this specific case, it is also an x-intercept, representing the maximum \( x \) value due to the leftward opening direction of the parabola.