Problem 11 - 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.. $$ y=-x^{2}-2 x+2 $$

Step-by-Step Solution

Answer

The graph is a downward-opening parabola with vertex (-1, 3), y-intercept (0, 2), and x-intercepts at approximately (2.73, 0) and (-0.73, 0).

Step 1: Identify the Equation Form

The given equation is \( y = -x^2 - 2x + 2 \). This is a quadratic equation and can be rewritten in the standard form: \( y = ax^2 + bx + c \), where \( a = -1 \), \( b = -2 \), and \( c = 2 \). Quadratic equations like this one represent parabolas.

Step 2: Determine the Symmetry

A parabola described by \( y = ax^2 + bx + c \) is symmetric about the vertical line \( x = -\frac{b}{2a} \). For this equation, calculate \( x = -\frac{-2}{2(-1)} = -1 \); thus, it is symmetric about the line \( x = -1 \).

Step 3: Find the Vertex

The vertex of a parabola given by a quadratic equation is at \( (h, k) \). Here, \( h = -1 \). To find \( k \), substitute \( x = -1 \) back into the equation: \( y = -(-1)^2 - 2(-1) + 2 = -1 + 2 + 2 = 3 \). So, the vertex is \( (-1, 3) \).

Step 4: Find the y-intercept

To find the \( y \)-intercept, set \( x = 0 \) and solve for \( y \): \( y = -(0)^2 - 2(0) + 2 = 2 \). Thus, the \( y \)-intercept is at \( (0, 2) \).

Step 5: Find the x-intercepts

Set \( y = 0 \) and solve the quadratic equation: \( 0 = -x^2 - 2x + 2 \). By the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = -1 \), \( b = -2 \), \( c = 2 \), find the roots: \( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(-1)(2)}}{2(-1)} \). Simplifying, \( x = \frac{2 \pm \sqrt{4 + 8}}{-2} = \frac{2 \pm \sqrt{12}}{-2} \). Further, \( x = \frac{2 \pm 2\sqrt{3}}{-2} \). Thus, the intercepts are \( x = 1 + \sqrt{3} \) and \( x = 1 - \sqrt{3} \).

Step 6: Plot and Sketch the Graph

Plot the vertex \((-1, 3)\), the \( y \)-intercept \((0, 2)\), and the \( x \)-intercepts \( (1+\sqrt{3}, 0) \) and \( (1-\sqrt{3}, 0) \). Draw a parabola opening downwards (since \( a = -1 \)). Ensure the symmetry line \( x = -1 \) is clearly depicted in the sketch.

Key Concepts

Parabola Symmetry

A quadratic equation like \( y = -x^2 - 2x + 2 \) graphically represents a parabola, which always shows a distinctive symmetry. This symmetry is about a vertical line known as the axis of symmetry. This line passes through the vertex of the parabola, meaning it evenly divides the parabola into two mirrored halves.
For a quadratic equation in the standard form \( y = ax^2 + bx + c \), you can locate the axis of symmetry with the formula \( x = -\frac{b}{2a} \). In our example, \( b = -2 \) and \( a = -1 \). Plug these values into the formula to find \( x = -\frac{-2}{2(-1)} = -1 \).
Thus, the parabola is symmetrical around the line \( x = -1 \). This line is crucial as it helps predict the behavior of the graph, making it easier to draw and understand parabolas.

Finding Intercepts

Intercepts are key points where the graph crosses the axes. Locating these intercepts helps understand the graph's interaction with the coordinate system. There are two types of intercepts:

• y-intercept: The point at which the graph crosses the y-axis. To find this intercept for the expression \( y = -x^2 - 2x + 2 \), set \( x = 0 \) and solve for \( y \). Substituting \( x = 0 \): \( y = -(0)^2 - 2(0) + 2 = 2 \). Therefore, the y-intercept is \( (0, 2) \).
• x-intercepts: These are points where the graph crosses the x-axis, occurring when \( y = 0 \). Solve \( 0 = -x^2 - 2x + 2 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here \( a = -1 \), \( b = -2 \), and \( c = 2 \), yielding:\[ x = \frac{2 \pm \sqrt{12}}{-2} = 1 \pm \sqrt{3} \].Therefore, the x-intercepts are \( (1 + \sqrt{3}, 0) \) and \( (1 - \sqrt{3}, 0) \).

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations, especially when finding x-intercepts. Quadratic equations have the form \( ax^2 + bx + c = 0 \). The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), provides the solution to this equation.
In our problem, the coefficients are \( a = -1 \), \( b = -2 \), and \( c = 2 \). Substitute these values into the formula to find the values of \( x \) that make \( y = 0 \). These are the x-intercepts of the equation:

• Calculate \( b^2 - 4ac \): \( (-2)^2 - 4(-1)(2) = 4 + 8 = 12 \).
• Substitute into the formula: \( x = \frac{2 \pm \sqrt{12}}{-2} \).
• Simplify: \( x = 1 \pm \sqrt{3} \).

The quadratic formula is not only useful for finding intercepts but also confirms the symmetric nature of parabolas, demonstrating how algebra links beautifully with graphing.

Vertex of a Parabola

The vertex is a vital point on a parabola, marking either the highest or lowest point. In the function \( y = -x^2 - 2x + 2 \), the vertex can be found using the same formula for the axis of symmetry, \( x = -\frac{b}{2a} \). From the calculations, \( x = -1 \). This value of \( x \) gives the horizontal position of the vertex.
To locate the vertex completely, substitute \( x = -1 \) back into the equation to find \( y \): \[y = -(-1)^2 - 2(-1) + 2 \]. Simplifying this gives:\[ y = -1 + 2 + 2 = 3 \]. Therefore, the vertex is \( (-1, 3) \).
The vertex provides insights into the direction the parabola opens. Since our \( a \) is negative, the parabola opens downwards, making the vertex the maximum point of the graph. Understanding the vertex aids in sketching the graph, illuminating its path and direction.