Problem 75 - Algebra and Trigonometry

Question

(a) On the same coordinate axes, sketch graphs (as accurately as possible) of the functions $$y=x^{3}-2 x^{2}-x+2 \quad \text { and } \quad y=-x^{2}+5 x+2$$ (b) On the basis of your sketch in part (a), at how many points do the two graphs appear to intersect? (c) Find the coordinates of all intersection points.

Step-by-Step Solution

Answer

The graphs intersect at three points: \((0, 2)\), \((3, 8)\), and \((-2, -12)\).

Step 1: Understand the Functions

We have two functions to graph: \( y = x^3 - 2x^2 - x + 2 \) and \( y = -x^2 + 5x + 2 \). The first is a cubic function, and the second is a quadratic function. We need to determine their key features to sketch them accurately.

Step 2: Analyze the Cubic Function

For the function \( y = x^3 - 2x^2 - x + 2 \):- **Intercepts**: - **Y-intercept**: Set \( x = 0 \), yielding \( y = 2 \). - **X-intercepts**: Solve \( x^3 - 2x^2 - x + 2 = 0 \) for real values.- **Behavior at Extremes**: As \( x \to \pm \infty \), the graph behaves like \( y = x^3 \), increasing at both ends.- **Critical Points**: Find the derivative \( y' = 3x^2 - 4x - 1 \) and solve \( y' = 0 \) to locate turning points.This gives the behavior of the graph.

Step 3: Analyze the Quadratic Function

For the function \( y = -x^2 + 5x + 2 \):- **Intercepts**: - **Y-intercept**: Setting \( x = 0 \) gives \( y = 2 \). - **X-intercepts**: Solve \( -x^2 + 5x + 2 = 0 \).- **Vertex**: The vertex form of a quadratic is found using \( x = -\frac{b}{2a} \), so here \( x = \frac{5}{-2 \times -1} = 2.5 \).The graph is a parabola opening downwards.

Step 4: Sketch the Graphs

Plot the intercepts, turning points, and vertex to construct the graphs. The cubic function will have a serpentine shape crossing the x-axis up to three times, while the quadratic will be an upside-down parabola that crosses the x-axis twice.

Step 5: Identify Intersection Points from Sketch

Based on the sketch, look for the number of points where the graphs intersect. Use the sketch to estimate these intersection coordinates. Typically, they intersect visually at two probable locations.

Step 6: Calculate Intersection Points Algebraically

Solve the equation \( x^3 - 2x^2 - x + 2 = -x^2 + 5x + 2 \) by simplifying to \( x^3 - x^2 - 6x = 0 \). Factor to find solutions:\[ x(x^2 - x - 6) = 0 \]Solve \( x = 0 \) and \( x^2 - x - 6 = 0 \), which factors to \( (x - 3)(x + 2) = 0 \), and find \( x = 3 \) or \( x = -2 \).

Step 7: Find Coordinates of Intersection Points

Substitute the x-values back into either original function to find corresponding y-values:- For \( x = 0 \), \( y = 2 \).- For \( x = 3 \), substitute into \( y = -x^2 + 5x + 2 \), giving \( y = -9 + 15 + 2 = 8 \).- For \( x = -2 \), substitute and find \( y = -4 - 10 + 2 = -12 \).Thus, intersection points are: \((0, 2), (3, 8), (-2, -12) \).

Key Concepts

Cubic Function

A cubic function is a polynomial function of degree three, and it can be expressed in the form \( y = ax^3 + bx^2 + cx + d \). Each of the coefficients \( a, b, c, \) and \( d \) are constants, and \( a eq 0 \). In this exercise, we have the function \( y = x^3 - 2x^2 - x + 2 \). This function's graph has distinct properties that are crucial for graph sketching.

• Intercepts: Find the y-intercept by setting \( x = 0 \), resulting in \( y = 2 \). For x-intercepts, solve the equation \( x^3 - 2x^2 - x + 2 = 0 \).
• Behavior at Extremes: As \( x \to \pm \infty \), the cubic function behaves similarly to \( y = x^3 \), demonstrating that the graph increases without bound at both extremes.
• Critical Points: To identify turning points, compute the derivative \( y' = 3x^2 - 4x - 1 \), and set it to zero to find values where the direction of the curve changes.

A cubic graph typically exhibits a serpentine shape, potentially intersecting the x-axis up to three times, depending on the roots of the equation.

Quadratic Function

Quadratic functions are polynomial functions of degree two and have the general form \( y = ax^2 + bx + c \). For this exercise, the quadratic is \( y = -x^2 + 5x + 2 \). Quadratics form parabolas that can open upwards or downwards depending on the sign of \( a \).

• Intercepts: The y-intercept for this function is \( y = 2 \) when \( x = 0 \). X-intercepts are found by setting \( y = 0 \), solving the quadratic \( -x^2 + 5x + 2 = 0 \).
• Vertex: The vertex is the highest or lowest point of the parabola, calculated as \( x = -\frac{b}{2a} \). For our equation, this is \( x = 2.5 \).

This function forms a downward-opening parabola, indicating that the vertex is a maximum point. The behavior of the graph is guided by these characteristics.

Intersection Points

Intersection points are locations on the graph where two functions have the same value of \( y \) for the same \( x \). In simple terms, they are where the graphs overlap. Finding these points involves setting the two function equations equal to each other and solving for \( x \).
To determine the intersection of \( y = x^3 - 2x^2 - x + 2 \) and \( y = -x^2 + 5x + 2 \), set the equations equal: \( x^3 - 2x^2 - x + 2 = -x^2 + 5x + 2 \). Simplifying gives \( x^3 - x^2 - 6x = 0 \).

• Factor and solve \( x(x^2 - x - 6) = 0 \).
• Solutions are \( x = 0 \), \( x = 3 \), or \( x = -2 \).

Substituting these \( x \)-values into either original equation yields the corresponding y-values, thus giving the intersection points: \((0, 2)\), \((3, 8)\), \((-2, -12)\). These points are crucial as they represent the values where both graphs meet.

Graph Sketching

Graph sketching involves plotting a function on a set of axes to visualize its behavior. For polynomial functions, like cubic and quadratic, understanding the traits like intercepts, critical points, and general shape assists in drawing accurate graphs. Graph sketching is a powerful tool that can reveal intersections and the overall flow of the functions.

• Identify intercepts and plot them on the axes. They provide preliminary points to begin sketching.
• Determine the direction and shape related to the leading term (for cubic, it can be serpentine; for quadratic, a parabola).
• Use turning points and vertex information to guide the pathway of the curve.

Ensure the essential features, such as correct steepness and orientation, match the functions' expectations. By sketching, the intersection points visually become evident, supporting both qualitative understanding and quantitative analysis of solutions.