Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 31-38, plot the graphs of both equations on the same coordinate plane. Find and label the points of intersection of the two graphs (see Example 4). $$ \begin{aligned} &y=-2 x+3 \\ &y=-2(x-4)^{2} \end{aligned} $$

Step-by-Step Solution

Answer

The graphs intersect at two points, calculated solutions involve irrational numbers related to the quadratic solution.

Step 1: Understand the equations and graph type

The equations given are both quadratic and linear. The first equation is linear: \( y = -2x + 3 \) which forms a straight line. The second equation is quadratic: \( y = -2(x-4)^2 \), representing a parabola. Our goal is to plot these on the same coordinate plane and find their intersection points.

Step 2: Plot the Linear Equation

Begin by plotting the linear equation \( y = -2x + 3 \). This line has a slope of -2 and a y-intercept of 3. Plot the y-intercept at (0, 3), then use the slope to find another point, such as (1, 1). Connect these points with a straight line.

Step 3: Plot the Quadratic Equation

Next, plot the quadratic equation \( y = -2(x-4)^2 \). This parabola opens downward (since it is multiplied by -2) and is centered at (4, 0). Since it's downward opening, choose points like (3, -2), (4, 0), and (5, -2) to sketch the parabola accurately.

Step 4: Determine Points of Intersection

To find the intersections, solve the equalities between the two equations. Set \( -2x + 3 = -2(x-4)^2 \) and solve for \( x \).Expand the quadratic equation: \( -2(x-4)^2 = -2(x^2 - 8x + 16) = -2x^2 + 16x - 32 \).Set equations to solve: \[ -2x + 3 = -2x^2 + 16x - 32 \]Rearrange:\[ 0 = -2x^2 + 18x - 35 \]Divide every term by -2 to simplify:\[ x^2 - 9x + rac{35}{2} = 0 \]Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = -9, c = \frac{35}{2} \).Calculate \( x \) values.

Step 5: Solve Quadratic Equation

Plug into the quadratic formula: \[ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \times 1 \times \frac{35}{2}}}{2 \times 1} \]Calculate the discriminant: \[ 81 - 70 = 11 \]Thus,\[ x = \frac{9 \pm \sqrt{11}}{2} \].Find the approximate \( x \) values using a calculator:

Step 6: Calculate y-values for Intersection

For each \( x \) solution, plug back into either equations (use \( y = -2x + 3 \)) to get \( y \)-coordinates. Upon solving, approximate intersections are:\[ (x_1, y_1) = \left( \frac{9 + \sqrt{11}}{2}, ext{ y value } \right) \] \[ (x_2, y_2) = \left( \frac{9 - \sqrt{11}}{2}, ext{ y value } \right) \].

Key Concepts

Linear Equation

Linear equations are mathematical expressions representing a straight line on the coordinate plane. A standard form for a linear equation is typically written as \( y = mx + b \), where \( m \) represents the slope, and \( b \) is the y-intercept.
The slope \( m \) indicates the steepness of the line and its direction. If \( m \) is positive, the line ascends from left to right; if negative, it descends. The y-intercept \( b \) is the point where the line crosses the y-axis.
In the given exercise, the linear equation is \( y = -2x + 3 \). Here, the slope is \(-2\), which means for each unit increase in \( x \), the \( y \) decreases by 2 units. The y-intercept is 3, indicating that the line crosses the y-axis at the point (0, 3). This straight line can be plotted by starting at the y-intercept and using the slope to determine another point, such as (1, 1). Connect these points with a neat straight line to complete its graph.

Quadratic Equation

A quadratic equation is a type of polynomial equation of the form \( ax^2 + bx + c = 0 \). It graphs as a parabola on the coordinate plane. The key feature of a quadratic equation is its curved shape.
The standard form is \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown. When a is positive, the parabola opens upward, resembling a cup; when \( a \) is negative, it opens downward.
In this problem, we have the quadratic equation \( y = -2(x-4)^2 \). The negative coefficient \(-2\) implies that the parabola opens downwards. The vertex form \((x-4)^2\) indicates that the parabola is shifted 4 units to the right on the x-axis and is centered at (4, 0). Points such as (3, -2), (4, 0), and (5, -2) can be used to sketch the parabola on the graph accurately.

Coordinate Plane

The coordinate plane is a two-dimensional surface where we plot mathematical functions like equations and inequalities.
This plane consists of two perpendicular axes:

• The horizontal axis, called the x-axis, where we plot x-values.
• The vertical axis, called the y-axis, where we plot y-values.

Each point on the plane is determined by a pair of numerical coordinates, which are typically written as (x, y). The origin, where the x-axis and y-axis intersect, is the point with coordinates (0, 0).
In our exercise, both the linear and the quadratic equations are represented on the same coordinate plane. This allows us to visually determine their intersection points - where both graphs cross each other on this plane.

Quadratic Formula

The quadratic formula is a mathematical formula used to find the solutions or "roots" of a quadratic equation.
The formula is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here:

• \( a \), \( b \), and \( c \) are constants from the equation \( ax^2 + bx + c = 0 \).
• \( \sqrt{b^2 - 4ac} \) is the discriminant, helping to determine the number and type of solutions.

In our solution, we simplify the intersection condition to \( x^2 - 9x + \frac{35}{2} = 0 \).
Using \( a = 1 \), \( b = -9 \), and \( c = 35/2 \), we apply the quadratic formula to find \( x \). The discriminant is calculated as 11, which reveals that there are two distinct, real solutions for x - given by the formula: \[ x = \frac{9 \pm \sqrt{11}}{2} \]. These solutions are used to find the intersection points along with their respective y-values from the linear equation.