Problem 45 - Algebra and Trigonometry

Question

The graph of a polynomial function is given. From the graph, find (a) the \(x\) - and \(y\) -intercepts (b) the coordinates of all local extrema $$ P(x)=-\frac{1}{2} x^{3}+\frac{3}{2} x-1 $$

Step-by-Step Solution

Answer

(a) x-intercept: (2,0); y-intercept: (0,-1) (b) Local max: (1,1); Local min: (-1,-1)

Step 1: Identify the x-intercepts

To find the \(x\)-intercepts, set \(P(x) = 0\) and solve for \(x\). This gives \(-\frac{1}{2}x^3 + \frac{3}{2}x - 1 = 0\). Solving this equation requires finding the roots, which can be efficiently done using graphical method, trial and error, or algebraic factorization. For visual verification, plot the graph to determine approximate \(x\)-intercepts. If \(x=2\) is a root, substituting gives the function equals zero, therefore, the \(x\)-intercept is \((x, y) = (2, 0)\).

Step 2: Identify the y-intercept

The \(y\)-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the function: \(P(x) = -\frac{1}{2}(0)^3 + \frac{3}{2}(0) - 1 = -1\). Thus, the \(y\)-intercept is \((0, -1)\).

Step 3: Find the derivative for local extrema

To find the local extrema, compute the first derivative of \(P(x)\): \(P'(x) = -\frac{3}{2}x^2 + \frac{3}{2}\). Set \(P'(x) = 0\) to find critical points: \(-\frac{3}{2}x^2 + \frac{3}{2} = 0\) simplifies to \(x^2 = 1\). Therefore, the critical points are \(x = 1\) and \(x = -1\).

Step 4: Determine the type of local extrema using second derivative

To determine whether these critical points are maxima or minima, evaluate the second derivative \(P''(x) = -3x\). Substitute \(x = 1\) and \(x = -1\) into \(P''(x)\): \(P''(1) = -3\cdot 1 = -3\) (which is less than 0, indicating a local maximum), and \(P''(-1) = -3\cdot (-1) = 3\) (which is greater than 0, indicating a local minimum).

Step 5: Calculate the coordinates of local extrema

From the critical points, calculate the function values: \(P(1) = -\frac{1}{2}(1)^3 + \frac{3}{2}(1) - 1 = 1\) for a local maximum, and \(P(-1) = -\frac{1}{2}(-1)^3 + \frac{3}{2}(-1) - 1 = -1\) for a local minimum. Thus, the local maximum is at \( (1, 1) \) and the local minimum is at \( (-1, -1) \).

Key Concepts

Understanding X-intercepts

An important feature of a polynomial graph is the x-intercepts. These points indicate where the graph crosses the x-axis. The value of the function at these points is zero. To find the x-intercepts mathematically, you set the polynomial equation equal to zero and solve for x. This involves solving the equation \[-\frac{1}{2}x^3 + \frac{3}{2}x - 1 = 0\]. In some instances, like when dealing with simple polynomials or those with rational solutions, these roots can be found via factoring or using graphical calculators for estimation. For our example polynomial, plotting the function reveals that one x-intercept is at x = 2, thus the coordinate is (2, 0). These intercepts are also referred to as the zeros or roots of the polynomial function.

Identifying Y-intercepts

Unlike x-intercepts, the y-intercept is the point where the graph intersects the y-axis. Here, the x-value is zero. To find the y-intercept of a polynomial function, substitute x = 0 into the polynomial equation and calculate the output. For the given polynomial function \(P(x) = -\frac{1}{2}x^3 + \frac{3}{2}x - 1\), we substitute x with 0 and simplify: \[-\frac{1}{2}(0)^3 + \frac{3}{2}(0) - 1 = -1\].Therefore, the y-intercept is at the point \( (0, -1) \). This point indicates the starting vertical position of the polynomial graph on the y-axis.

Discovering Local Extrema

Local extrema describe the high or low points on a graph that are not necessarily the highest or lowest overall but are relatively so compared to nearby points. These local extremes occur at critical points.To find these points for a polynomial function, you calculate the derivative and set it to zero. For our polynomial,\[P'(x) = -\frac{3}{2}x^2 + \frac{3}{2}\]. Setting this derivative equal to zero \(-\frac{3}{2}x^2 + \frac{3}{2} = 0\) provides values of x where the graph direction changes, hence x = 1 and x = -1 are the critical points.To classify these points as maxima or minima, the second derivative \(P''(x) = -3x\) is used. A negative value indicates a local maximum, while a positive value indicates a local minimum. Here, \(x = 1\) results in a local maximum at \( (1, 1) \) and \(x = -1\) yields a local minimum at \( (-1, -1) \).

Understanding Critical Points

Critical points are key to locating changes in the direction of a graph, often leading to local extrema. They occur where the first derivative is zero or undefined. In polynomial functions, derivatives use power rules and can be set up straightforwardly.For the function \(P(x) = -\frac{1}{2}x^3 + \frac{3}{2}x - 1\), the first derivative, \[P'(x) = -\frac{3}{2}x^2 + \frac{3}{2}\], is key in identifying changes in slope direction. By solving \(P'(x) = 0\), we find x = 1 and x = -1.These critical x-values allow us to plug back into the original function to verify function values: - At \( x = 1 \): \( P(1) = 1 \), marking a local maximum. - At \( x = -1 \): \( P(-1) = -1 \), indicating a local minimum.Understanding and identifying these points help in comprehending the overall shape and nature of polynomial curves. This enhances problem solving, including optimization and modeling scenarios.