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Chapter 4: Problem 51

Give a graph of the polynomial and label the coordinates of the intercepts, stationary points, and inflection points. Check your work with a graphing utility. $$ p(x)=x^{2}-3 x-4 $$

Short Answer

Expert verified
The intercepts are (0,-4), (4,0), (-1,0) and the stationary point is (3/2, -25/4). There are no inflection points.

Step by step solution

01

Find the y-intercept

To find the y-intercept of the polynomial, set \( x = 0 \). Calculate \( p(0) = 0^2 - 3(0) - 4 = -4 \). Therefore, the y-intercept is at the point \((0, -4)\).
02

Find the x-intercepts

Find the roots of \( p(x) = 0 \) by solving the equation \( x^2 - 3x - 4 = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -3 \), and \( c = -4 \). Calculate:\[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} \]This gives us the solutions: \( x_1 = 4 \) and \( x_2 = -1 \). Thus, the x-intercepts are \((4, 0)\) and \((-1, 0)\).
03

Find the stationary points

First, compute the derivative of the polynomial, \( p'(x) = 2x - 3 \). Set the derivative to zero to find the stationary points: \( 2x - 3 = 0 \) leads to \( x = \frac{3}{2} \). Evaluate the original polynomial at this point: \( p\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) - 4 = -\frac{25}{4} \). Therefore, the stationary point is \( \left(\frac{3}{2}, -\frac{25}{4}\right) \).
04

Check for inflection points

Find the second derivative of the polynomial, which is \( p''(x) = 2 \). Since the second derivative is constant and not equal to zero, there are no inflection points for this polynomial curve.
05

Sketch the graph

Using the intercepts and the stationary point, sketch the polynomial curve. The parabola opens upwards as the coefficient of \( x^2 \) is positive. Ensure that the points \((0, -4)\), \((4, 0)\), \((-1, 0)\), and \(\left(\frac{3}{2}, -\frac{25}{4}\right)\) are marked accurately on the graph. Since there is no change in concavity, the curve will have a consistent U-shape.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Y-intercept Calculation
Finding the y-intercept of a polynomial is straightforward. You only need to set the value of x to zero and calculate the resulting y-value. This gives you the point where the graph crosses the y-axis.

Here's how it works for our given polynomial, \( p(x) = x^2 - 3x - 4 \):
  • Set \( x = 0 \) in the polynomial expression.
  • Calculate \( p(0) = 0^2 - 3(0) - 4 = -4 \).
The y-intercept is therefore located at the point \((0, -4)\).

This means that the graph will intersect the y-axis at \( -4 \), giving you a solid starting point for sketching your graph.
X-intercepts Finding
X-intercepts are the points where the graph of the polynomial crosses the x-axis. These occur where the value of the polynomial is zero, meaning the y-coordinate is 0. To find x-intercepts, set the polynomial equation equal to zero and solve for x. This often involves using the quadratic formula.

Let's use our polynomial for illustration: \( x^2 - 3x - 4 = 0 \).

Applying the quadratic formula:
  • The formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
  • For our equation, \( a = 1 \), \( b = -3 \), and \( c = -4 \).
  • Calculate:
    \[ x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)} = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2} \]
  • This yields two values: \( x_1 = 4 \) and \( x_2 = -1 \).
So, the x-intercepts are at the points \((4, 0)\) and \((-1, 0)\).

These are crucial for plotting because they show where the graph touches the x-axis.
Stationary Points
Stationary points are where the slope of the graph (or the derivative) is zero. These points are essential for understanding the shape of the graph, such as peaks or troughs.

For the polynomial \( p(x) = x^2 - 3x - 4 \), find the derivative, \( p'(x) = 2x - 3 \).
  • Set \( p'(x) = 0 \) to find the stationary points.
    \( 2x - 3 = 0 \) gives \( x = \frac{3}{2} \).
  • Evaluate this x-value in the original polynomial:
    \( p\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) - 4 = -\frac{25}{4} \).
This calculates to the stationary point \( \left(\frac{3}{2}, -\frac{25}{4}\right) \).

Locating this point helps in sketching because it shows where the curve of the graph changes direction or flattens out.
Quadratic Formula
The quadratic formula is a powerful tool for finding the roots of a quadratic equation, which are the x-intercepts of a polynomial graph.

The general form of the formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula helps you solve any quadratic equation of the form \( ax^2 + bx + c = 0 \).
  • Identify coefficients \( a \), \( b \), and \( c \) from the polynomial expression.
  • Substitute these values into the formula to find the values of x.
  • The two solutions from the formula represent the x-intercepts of the graph.
This method is reliable and ensures you find the correct intercepts, forming the foundation for accurate polynomial graphing.

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Most popular questions from this chapter

The accompanying figure shows the path of a fly whose equations of motion are $$x=\frac{\cos t}{2+\sin t}, \quad y=3+\sin (2 t)-2 \sin ^{2} t \quad(0 \leq t \leq 2 \pi)$$ (a) How high and low does it fly? (b) How far left and right of the origin does it fly?

Sketch a reasonable graph of \(s\) versus \(t\) for a mouse that is trapped in a narrow corridor (an \(s\) -axis with the positive direction to the right) and scurries back and forth as follows. It runs right with a constant speek and forth as for a while, then gradually slows down \(t 0.6 \mathrm{m} / \mathrm{s}\), then quickly speeds up to \(2.0 \mathrm{m} / \mathrm{s}\), then gradually slows to a stop but immediately reverses direction and quickly speeds up to \(1.2 \mathrm{m} / \mathrm{s}\).

The function \(s(t)\) describes the position of a particle moving along a coordinate line, where \(s\) is in feet and \(t\) is in seconds. (a) Find the velocity and acceleration functions. (b) Find the position, velocity, speced, and acceleration at time \(t=1 .\) (c) At what times is the particle stopped? (d) When is the particle speeding up? Slowing down? (e) Find the total distance traveled by the particle from time \(t=0\) to time \(t=5\). $$ s(t)=\frac{t}{t^{2}+4}, \quad t \geq 0 $$

Verify that the hypotheses of Rolle's Theorem are satisfied on the given interval, and find all values of \(c\) in that interval that satisfy the conclusion of the theorem. $$ f(x)=x^{2}-8 x+15 ;[3,5] $$

Writing Discuss the relative advantages or disadvantages of using the first derivative test versus using the second derivative test to classify candidates for relative extrema on the interior of the domain of a function. Include specific examples to illustrate your points.
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