Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

plot the graphs of both equations on the same coordinate plane. Find and label the points of intersection of the two graphs. $$ \begin{array}{l} y=-2 x+3 \\ y=-2(x-4)^{2} \end{array} $$

Step-by-Step Solution

Answer

The intersection points of the plots are solutions to \(-2x^2 + 18x - 35 = 0\). Solve for \(x\), then find corresponding \(y\)-values using the linear equation.

Step 1: Understand the Equations

First equation is a linear equation: \(y = -2x + 3\). The second equation is a quadratic equation: \(y = -2(x-4)^2\). Understanding these forms helps in knowing that one is a straight line and the other is a parabola.

Step 2: Determine Key Coordinates

For the linear equation \(y = -2x + 3\), find the y-intercept (where \(x=0\)) to get \(y=3\), and the x-intercept (where \(y=0\)) to solve \(-2x + 3 = 0\). For the quadratic equation \(y = -2(x-4)^2\), the vertex is at \((4,0)\), since the vertex form is \(y = a(x-h)^2 + k\).

Step 3: Plot Critical Points on Graph

Plot the points for the linear equation: the intercepts (0,3) and (1.5,0). For \(y = -2(x-4)^2\), plot the vertex (4,0) and some additional points, for example, calculate \(y\) for \(x=3\) and \(x=5\).

Step 4: Draw Graphs for Each Equation

Draw the line for \(y = -2x + 3\), connecting the calculated points. Draw the parabola for \(y = -2(x-4)^2\) that opens downwards, symmetric around \(x=4\).

Step 5: Solve for Intersection Points

Set the equations equal to find the intersection points algebraically: \(-2x + 3 = -2(x-4)^2\). Expand the quadratic: \(-2(x-4)^2 = -2(x^2 - 8x + 16) = -2x^2 + 16x - 32\). Equate and simplify: \(-2x + 3 = -2x^2 + 16x - 32\). Rearrange to: \(-2x^2 + 18x - 35 = 0\).

Step 6: Find Roots of Quadratic

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find roots of \(-2x^2 + 18x - 35 = 0\). Here, \(a=-2\), \(b=18\), \(c=-35\). Substitute values to find roots.

Step 7: Substitute to Find Intersection Points

Once roots (x-values) are found, substitute back into the original linear equation \(y = -2x + 3\) to find y-values for each x intersection. Label these points on the graph.

Key Concepts

Linear Equations

Linear equations are a fundamental concept in mathematics with a myriad of applications in various fields. A linear equation is an equation between two variables that produces a straight line when graphed. It typically takes the form \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.

• Slope \(m\): Indicates the steepness or incline of the line. A positive slope means the line rises as it moves from left to right, while a negative slope means the opposite. In our example, the slope is \(-2\), indicating a downward slope.
• Y-Intercept \(b\): The point where the line crosses the y-axis. It occurs when \(x = 0\). For our equation \(y = -2x + 3\), the y-intercept is \(3\).

In practice, we use these components to draw the line on a graph, starting with the y-intercept and using the slope to determine the direction and angle of the line.

Quadratic Equations

Quadratic equations are a key topic in algebra and are characterized by their U-shaped graph called a parabola. The standard form of a quadratic equation is \(y = ax^2 + bx + c\). However, the vertex form \(y = a(x-h)^2 + k\) is often used when graphing quadratics.

• Vertex: The highest or lowest point of the parabola, given by \((h,k)\) in the vertex form. In our example, the equation is \(y = -2(x-4)^2\), so the vertex is at \((4, 0)\).
• Direction: The coefficient \(a\) affects the direction of the parabola. If \(a\) is positive, the parabola opens upwards; if negative, it opens downwards. Here, \(a = -2\), thus opening downwards.

Additional points can be found by choosing x-values around the vertex and calculating the corresponding y-values. This helps give a complete picture of the parabola on the coordinate plane.

Points of Intersection

Points of intersection between two graphs indicate where the graphs meet or cross each other. Finding these points involves setting the equations equal to each other and solving for the values of the variables.

• Set Equations Equal: For our equations, \(-2x + 3 = -2(x-4)^2\), we expand and simplify the quadratic equation to solve for \(x\).
• Algebraic Solution: Rearranging gives \(-2x^2 + 18x - 35 = 0\). Applying the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we determine the x-coordinates of the intersection.
• Substitute Back: Once \(x\) is found, substitute it back into one of the original equations, such as the linear equation \(y = -2x + 3\), to find the corresponding y-values.

These steps help us pinpoint exactly where the linear and quadratic graph intersect on the coordinate plane.

Coordinate Plane

The coordinate plane is a two-dimensional plane formed by the intersection of two perpendicular number lines: the x-axis (horizontal) and the y-axis (vertical). It is essential for graphing equations because it provides a visual way to analyze mathematical relationships.

• Axes: The horizontal line is the x-axis, and the vertical line is the y-axis. Each point on the plane is represented by an ordered pair \((x, y)\).
• Quadrants: The plane is divided into four quadrants. Quadrant I has positive x and y values; Quadrant II has negative x and positive y; Quadrant III has both negative x and y; and Quadrant IV has positive x and negative y.

On the coordinate plane, you can accurately plot points and draw graphs of linear and quadratic equations to find the points of intersection. This makes it easier to comprehend the relationship between the equations physically and visually.