Problem 26 - 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.. $$ y=x^{2}(x-1)(x-2) $$

Step-by-Step Solution

Answer

The graph of \( y = x^2(x-1)(x-2) \) intersects the x-axis at (0,0), (1,0), and (2,0) with no symmetries and rises at the ends.

Step 1: Check Symmetries

To check for symmetries, first test for y-axis symmetry by replacing \( x \) with \( -x \) in the equation.\[ y = (-x)^2(-x-1)(-x-2) = x^2(x+1)(x+2) \] This is not the same as the original equation, so there is no y-axis symmetry. Next, check for origin symmetry by replacing both \( x \) and \( y \) with \( -x \) and \( -y \) respectively. \[ -y = x^2(x+1)(x+2) \] Again, this is not the same as the original, so there is no origin symmetry. Hence, the graph is not symmetric about the y-axis or the origin.

Step 2: Find the x-intercepts

To find the x-intercepts, set \( y = 0 \) and solve for \( x \). The equation becomes \( x^2(x-1)(x-2) = 0 \). Setting each factor equal to zero gives the x-intercepts: \( x = 0, 1, \) and \( 2 \). Therefore, the graph intersects the x-axis at these points.

Step 3: Find the y-intercept

To find the y-intercept, set \( x = 0 \) in the equation and solve for \( y \). This gives \( y = 0^2(0-1)(0-2) = 0 \), so the y-intercept is \( 0 \).

Step 4: Analyze the End Behavior

Analyze the behavior of the graph as \( x \to \pm \infty \). The function is a polynomial of degree 4, so as \( x \to \infty \) or \( x \to -\infty \), \( y \to \infty \). This indicates that the graph will rise in both directions at the ends.

Step 5: Sketch the Graph

Begin plotting the graph using the information gathered. The points (0,0), (1,0), and (2,0) are x-intercepts; the graph crosses the x-axis at these points. The y-intercept is at (0,0). Since there is no symmetry, plot additional points to capture the shape of the curve between the intercepts. Consider the end behavior showing the rising curves as \( x \to \pm \infty \). The graph will have turns due to it being a degree 4 polynomial.

Key Concepts

Symmetry in Graphs

When analyzing a polynomial graph, understanding symmetry helps simplify graphing efforts and predicts certain behaviors. Let's break this down:

• Y-axis Symmetry: If a graph is symmetric about the y-axis, substituting \( x \) with \( -x \) in the equation should yield the original equation. Here, we substitute \( x = -x \) in \( y = x^2(x-1)(x-2) \), leading to a new equation, \( y = x^2(x+1)(x+2) \), which differs from the original. Thus, no y-axis symmetry exists.
• Origin Symmetry: For origin symmetry, change both the signs of \( x \) and \( y \) (i.e., replace \( y \) with \( -y \), and \( x \) with \( -x \)). This leads to \( -y = (-x)^2(-x-1)(-x-2) = x^2(x+1)(x+2) \), which again doesn't match the original equation. Hence, no origin symmetry is present.

By checking for these symmetries, we eliminate possibilities, confirming the graph is neither y-axis nor origin symmetric. This clarity helps in the accurate rendering of the polynomial's curve.

X-intercepts

X-intercepts are points where the graph crosses the x-axis, occurring where the value of \( y \) is zero. To find these intercepts in our polynomial \( y = x^2(x-1)(x-2) \), we solve:

• Set \( y = 0 \), resulting in the equation \( x^2(x-1)(x-2) = 0 \).
• By setting each factor equal to zero, we get the roots: \( x = 0 \), \( x = 1 \), and \( x = 2 \).

These x-values indicate where the graph touches or crosses the x-axis: at points (0,0), (1,0), and (2,0). Understanding x-intercepts is crucial for sketching as they anchor the curve’s intersections with the x-axis.

Y-intercepts

Y-intercepts occur where the graph crosses the y-axis, meaning \( x = 0 \). Let's determine our graph's y-intercept:

• Substitute \( x = 0 \) into the equation: \( y = 0^2(0-1)(0-2) \). Simplifying, \( y = 0 \).

Thus, the y-intercept is (0,0). This aligns with one of our x-intercepts, highlighting where the graph intersects both the x and y axes. It is always a good practice to find the y-intercept as it helps further pinpoint the graph’s starting point on the y-axis, aiding in a more accurate sketch.

End Behavior of Polynomials

In polynomial graphing, understanding the end behavior is vital as it describes how the graph behaves as \( x \) approaches positive or negative infinity. The end behavior largely depends on the degree and leading coefficient of the polynomial:

• Our polynomial is of degree 4 \((x^4)\), an even power, indicating that as \( x \to \pm \infty \), both ends of the graph rise \( y \to \infty \).
• This occurs because in even-degree polynomials, the leading coefficient being positive causes both ends of the graph to point upwards.

Understanding this helps in sketching, as you visualize both arms of the polynomial's curve heading upwards, giving insight into the long-range layout of your graph.