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

Step-by-Step Solution

Answer

The graph is a downward-opening parabola with vertex at \((0, 1)\), x-intercepts at \((1, 0)\) and \((-1, 0)\), and y-intercept at \((0, 1)\).

Step 1: Identify the Type of Equation

The given equation is \( y = -x^2 + 1 \), which is a quadratic equation. This means the graph will be a parabola.

Step 2: Check for Symmetries

To check for symmetries, analyze the equation. Since the equation is in the form \( y = -x^2 + 1 \), it is symmetric about the y-axis. This is because replacing \( x \) with \( -x \) results in the same equation.

Step 3: Find the Vertex

For a parabola given by \( y = ax^2 + bx + c \), the vertex can be calculated using \( x = -\frac{b}{2a} \). Here, \( a = -1 \), \( b = 0 \), and \( c = 1 \), so \( x = 0 \). Substitute this back into the equation to find \( y = 1 \). Thus, the vertex is \((0, 1)\).

Step 4: Find the Axis of Symmetry

The axis of symmetry is the vertical line that passes through the vertex. For this equation, it is \( x = 0 \).

Step 5: Find the x-Intercepts

To find the x-intercepts, set \( y = 0 \) and solve for \( x \): \(-x^2 + 1 = 0\). Rearrange this to \( x^2 = 1 \). Solving for \( x \), we get \( x = \pm 1 \). Thus, the x-intercepts are \((1, 0)\) and \((-1, 0)\).

Step 6: Find the y-Intercept

The y-intercept occurs where \( x = 0 \). Substitute \( x = 0 \) into the equation: \( y = -0^2 + 1 = 1 \). Therefore, the y-intercept is \((0, 1)\).

Step 7: Plot the Parabola

Plot the points found: the vertex \((0, 1)\), the x-intercepts \((1, 0)\) and \((-1, 0)\), and the y-intercept \((0, 1)\). Draw the parabola opening downwards as the term \(-x^2\) indicates the coefficient is negative.

Key Concepts

Symmetry in Graphs

Understanding symmetry in graphs is crucial for analyzing and drawing quadratic equations. Symmetry refers to the balance or mirror image in a graph when divided by a specific line. For quadratic equations, like our example \( y = -x^2 + 1 \), symmetry is often observed along the y-axis. This is because the equation remains unchanged when replacing \( x \) with \( -x \).

• This y-axis symmetry simplifies plotting as each point on one side of the y-axis has a mirrored point on the other side, making the graph predictable and easier to draw.
• Because symmetry confirms that the vertex lies on the axis of symmetry, it is a powerful tool in confirming the parabola's shape and position.

Checking for symmetry helps reduce the work needed in graph plotting, as only half of the parabola needs detailed checking.

Parabola Vertex

The vertex of a parabola is its highest or lowest point, depending on the parabola's direction. For the equation \( y = -x^2 + 1 \), the parabola's vertex is at \((0, 1)\).
To find the vertex, use the formula \( x = -\frac{b}{2a} \). In the standard quadratic form \( y = ax^2 + bx + c \), identify:

• \( a = -1 \)
• \( b = 0 \)
• \( c = 1 \)

Substitute into the vertex formula: \( x = -\frac{0}{2(-1)} = 0\). Plugging \( x = 0 \) back into the original equation gives the y-value \( y = 1 \). The vertex \((0, 1)\) is essential as it provides both the direction of the parabola (shown by the sign of \( a \)) and the maximum or minimum point on the graph.

X-intercepts

X-intercepts are the points where the graph crosses the x-axis. At these points, \( y = 0 \). For our equation \( y = -x^2 + 1 \), to find the x-intercepts, set the equation to zero:
\(-x^2 + 1 = 0\). Rearrange to solve for \( x^2 = 1 \), giving us \( x = \pm 1 \).

• These values mean the parabola crosses the x-axis at \((1, 0)\) and \((-1, 0)\).
• Knowing these intercepts gives us crucial points that define the parabola's interaction with the x-axis.

X-intercepts are significant because they show us where the function reaches a value of zero. This calculating step also provides valuable insight for solving real-world problems involving maximum or minimum values in quadratic scenarios.

Y-intercepts

The y-intercept is where a graph intersects the y-axis. For any function, the y-intercept can be found by evaluating the function when \( x = 0 \).
For our equation \( y = -x^2 + 1 \), substitute \( x = 0 \):
\( y = -0^2 + 1 = 1 \). Therefore, the y-intercept is \((0, 1)\).

• This is the same point as the vertex due to the parabola's symmetry.
• For quadratic graphs, particularly those symmetric about the y-axis, the y-intercept often coincides with the vertex's y-coordinate.

Understanding y-intercepts is foundational when sketching and understanding graphs, as they provide a fixed point on the y-axis around which the rest of the graph can be oriented. It gives a starting point for plotting and further analysis.