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 x-intercepts are at (4, 0) and (-1, 0), the y-intercept is at (0, -4), and the stationary point is at (3/2, -25/4). There are no inflection points.
Step by step solution
01
Find the x-intercepts
The x-intercepts occur where the polynomial equals zero. Solve for \(x\) in 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 \(c = -4\). Substituting we get \(x = \frac{3 \pm \sqrt{9 + 16}}{2}\), resulting in \(x = \frac{3 \pm 5}{2}\). Thus, the x-intercepts are \(x = 4\) and \(x = -1\).
02
Find the y-intercept
The y-intercept is found by evaluating \(p(x)\) at \(x = 0\). Substitute \(x = 0\) into the polynomial: \(p(0) = 0^2 - 3(0) - 4 = -4\). The y-intercept is at the point \((0, -4)\).
03
Find the stationary points
Stationary points are found by setting the derivative \(p'(x)\) to zero. First, find the derivative: \(p'(x) = 2x - 3\). Set \(2x - 3 = 0\), solving gives \(x = \frac{3}{2}\). Evaluate \(p\left(\frac{3}{2}\right)\) to get the y-coordinate: \(p\left(\frac{3}{2}\right) = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) - 4 = -\frac{25}{4}\). The stationary point is \(\left(\frac{3}{2}, -\frac{25}{4}\right)\).
04
Determine the inflection points
A quadratic polynomial does not have an inflection point. Inflection points occur where the concavity of the graph changes, but since the second derivative \(p''(x) = 2\) is constant and doesn't change sign, there are no inflection points.
05
Sketch the graph
Using the x-intercepts \((4, 0)\) and \((-1, 0)\), y-intercept \((0, -4)\), and stationary point \(\left(\frac{3}{2}, -\frac{25}{4}\right)\), sketch the graph. Mark these important points and draw a parabola opening upwards. Verify the graph using a graphing utility.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a crucial tool when dealing with polynomials, especially quadratics. It provides a simple way to find the roots of any quadratic equation in the standard form of \(ax^2 + bx + c = 0\). The formula itself is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula derives from completing the square on the generic quadratic equation.
- **a, b, c**: These are constants where \(aeq 0\). - **Discriminant**: The part under the square root symbol \(b^2 - 4ac\) is called the discriminant. It can tell us about the nature of the roots:
- If it's positive, there are two distinct real roots. - If it's zero, there is one real root (a repeated root). - If it's negative, the roots are complex numbers.
By using this formula, we found the roots of the polynomial \(p(x) = x^2 - 3x - 4\) to be \(x = 4\) and \(x = -1\). These roots correspond to the x-intercepts of the graph.
- **a, b, c**: These are constants where \(aeq 0\). - **Discriminant**: The part under the square root symbol \(b^2 - 4ac\) is called the discriminant. It can tell us about the nature of the roots:
- If it's positive, there are two distinct real roots. - If it's zero, there is one real root (a repeated root). - If it's negative, the roots are complex numbers.
By using this formula, we found the roots of the polynomial \(p(x) = x^2 - 3x - 4\) to be \(x = 4\) and \(x = -1\). These roots correspond to the x-intercepts of the graph.
X-intercepts
X-intercepts are points where the graph of a function crosses the x-axis. This means the y-value at these points is zero. For a quadratic function, the x-intercepts are also known as the roots or solutions of the equation.
To find these points for the function \(p(x) = x^2 - 3x - 4\), we solved the equation \(x^2 - 3x - 4 = 0\) using the quadratic formula, leading to two solutions: \(x = 4\) and \(x = -1\). These solutions tell us where the graph of the polynomial touches or crosses the x-axis, resulting in the intercepts at points \((4, 0)\) and \((-1, 0)\).
It's important to graph these points accurately as they help in shaping the curve of the parabola accurately.
To find these points for the function \(p(x) = x^2 - 3x - 4\), we solved the equation \(x^2 - 3x - 4 = 0\) using the quadratic formula, leading to two solutions: \(x = 4\) and \(x = -1\). These solutions tell us where the graph of the polynomial touches or crosses the x-axis, resulting in the intercepts at points \((4, 0)\) and \((-1, 0)\).
It's important to graph these points accurately as they help in shaping the curve of the parabola accurately.
Stationary Points
A stationary point on a graph occurs where the derivative of a function equals zero. At this point, the graph has a horizontal tangent and represents a local maximum, minimum, or a saddle point. For the polynomial \(p(x) = x^2 - 3x - 4\), we found the derivative to be \(p'(x) = 2x - 3\).
Setting the derivative equal to zero, \(2x - 3 = 0\), we solved for \(x\) and found \(x = \frac{3}{2}\). Evaluating the polynomial at this point gives us \(p\left(\frac{3}{2}\right) = -\frac{25}{4}\), so the stationary point is \(\left(\frac{3}{2}, -\frac{25}{4}\right)\).
This point is essential when sketching the graph, as it provides information about the turning point of the curve, helping to define whether the point is a minimum or maximum. In this case, since the parabola opens upwards, \(\left(\frac{3}{2}, -\frac{25}{4}\right)\) is a local minimum.
Setting the derivative equal to zero, \(2x - 3 = 0\), we solved for \(x\) and found \(x = \frac{3}{2}\). Evaluating the polynomial at this point gives us \(p\left(\frac{3}{2}\right) = -\frac{25}{4}\), so the stationary point is \(\left(\frac{3}{2}, -\frac{25}{4}\right)\).
This point is essential when sketching the graph, as it provides information about the turning point of the curve, helping to define whether the point is a minimum or maximum. In this case, since the parabola opens upwards, \(\left(\frac{3}{2}, -\frac{25}{4}\right)\) is a local minimum.
Inflection Points
While studying quadratic polynomials, it's important to note the absence of inflection points. Inflection points are where a curve changes its concavity, from concave up to concave down, or vice versa.
The presence of inflection points is determined by the second derivative of a function. If the second derivative changes its sign at a particular point, then that point is an inflection point. For the polynomial \(p(x) = x^2 - 3x - 4\), we find that the second derivative is \(p''(x) = 2\), a constant.
Since the second derivative does not change sign, there are no inflection points for this polynomial. This tells us that the graph of a quadratic function is entirely concave up or concave down, never changing in between, making it a smooth, symmetrical curve.
The presence of inflection points is determined by the second derivative of a function. If the second derivative changes its sign at a particular point, then that point is an inflection point. For the polynomial \(p(x) = x^2 - 3x - 4\), we find that the second derivative is \(p''(x) = 2\), a constant.
Since the second derivative does not change sign, there are no inflection points for this polynomial. This tells us that the graph of a quadratic function is entirely concave up or concave down, never changing in between, making it a smooth, symmetrical curve.