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Chapter 4: Problem 4

For each quadratic function, state whether it would make sense to look for a highest or a lowest point on the graph. Then determine the coordinates of that point. (a) \(y=2 x^{2}-8 x+1\) (b) \(y=-3 x^{2}-4 x-9\) (c) \(h=-16 t^{2}+256 t\) (d) \(f(x)=1-(x+1)^{2}\) (e) \(g(t)=t^{2}+1\) (f) \(f(x)=1000 x^{2}-x+100\)

Short Answer

Expert verified
Find the lowest point for (a), (e), (f) and highest for (b), (c), (d) with respective vertices (2, -7), \((\frac{2}{3}, -\frac{37}{3})\), (8, 1024), (-1, 1), (0, 1), \((\frac{1}{2000}, 99.9995)\).

Step by step solution

01

Determine the vertex for (a)

The quadratic function in (a) is \( y = 2x^2 - 8x + 1 \). This parabola opens upwards because the leading coefficient (2) is positive. Therefore, we are interested in finding the lowest point, which is the vertex. The x-coordinate of the vertex can be found using the formula \( x = -\frac{b}{2a} \), where \( a = 2 \) and \( b = -8 \).\[ x = -\frac{-8}{2 \cdot 2} = 2 \]Plug \( x = 2 \) back into the polynomial to find the y-coordinate.\[ y = 2(2)^2 - 8(2) + 1 = -7 \]Thus, the coordinates of the vertex are \((2, -7)\).
02

Determine the vertex for (b)

The quadratic function in (b) is \( y = -3x^2 - 4x - 9 \). This parabola opens downwards because the leading coefficient (-3) is negative. Hence, we are interested in finding the highest point, which is the vertex. The x-coordinate of the vertex is:\[ x = -\frac{-4}{2 \cdot (-3)} = \frac{2}{3} \]Plug \( x = \frac{2}{3} \) into the polynomial to find the y-coordinate:\[ y = -3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) - 9 = -\frac{37}{3} \]Thus, the coordinates of the vertex are \(\left(\frac{2}{3}, -\frac{37}{3}\right)\).
03

Determine the vertex for (c)

For the function \( h = -16t^2 + 256t \), the parabola opens downwards because the leading coefficient (-16) is negative. Thus, we need the highest point. The x-coordinate of the vertex is:\[ t = -\frac{256}{2 \times (-16)} = 8 \]Plug \( t = 8 \) back to find the h-coordinate:\[ h = -16(8)^2 + 256(8) = 1024 \]The coordinates of the vertex are \((8, 1024)\).
04

Determine the vertex for (d)

The function is \( f(x) = 1 - (x+1)^2 \). Expand it and identify the form:\[ f(x) = 1 - (x^2 + 2x + 1) = -x^2 - 2x + 1 \]The parabola opens downwards (leading coefficient -1) indicating a maximum point at the vertex. Using the vertex formula:\[ x = -\frac{-2}{2 \cdot (-1)} = -1 \]Substitute \( x = -1 \) to find the y-coordinate:\[ f(-1) = 1 - (-1+1)^2 = 1 \]The vertex is \((-1, 1)\).
05

Determine the vertex for (e)

For \( g(t) = t^2 + 1 \), the parabola opens upwards because the leading coefficient (1) is positive, so we find the lowest point. The x-coordinate of the vertex is:\[ t = -\frac{0}{2 \cdot 1} = 0 \]Substitute \( t = 0 \) back into the function to find the y-coordinate:\[ g(0) = (0)^2 + 1 = 1 \]The vertex's coordinates are \((0, 1)\).
06

Determine the vertex for (f)

The quadratic function is \( f(x) = 1000x^2 - x + 100 \), opening upwards since the leading coefficient (1000) is positive. So, we find the lowest point. Using the vertex formula:\[ x = -\frac{-1}{2 \cdot 1000} = \frac{1}{2000} \]Substitute \( x = \frac{1}{2000} \) into the polynomial:\[ f\left(\frac{1}{2000}\right) = 1000\left(\frac{1}{2000}\right)^2 - \frac{1}{2000} + 100 \approx 99.9995 \]The vertex is approximately \(\left(\frac{1}{2000}, 99.9995\right)\).

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Key Concepts

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

Parabola
A parabola is the graph of a quadratic function, and it has a unique curved shape. Its general equation can be expressed as \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. The shape and direction of the parabola are determined by the value of the coefficient \( a \).
A parabola can open upwards or downwards:
  • If \( a > 0 \) (positive), the parabola opens upwards, creating a U-shape.
  • If \( a < 0 \) (negative), the parabola opens downwards, forming an inverted U-shape.
The point where the direction of the curve changes is called the vertex, which can either be a minimum or maximum point depending on the parabola's direction.
Vertex of Quadratic
The vertex of a quadratic function is a significant point on the graph as it provides valuable information about the function's values. To find the vertex, we use the formula \( x = -\frac{b}{2a} \). Once the x-coordinate is determined, it can be substituted back into the quadratic equation to calculate the y-coordinate.
Here are the steps to find the vertex:
  • Identify the coefficients \( a \) and \( b \) from the equation.
  • Apply the formula \( x = -\frac{b}{2a} \) to calculate the x-coordinate.
  • Plug the x-coordinate back into the equation to find the y-coordinate.
The vertex is expressed as a coordinate pair \( (x, y) \). This point is where the parabola achieves either its maximum height or lowest depth.
Graphing Quadratics
Graphing quadratics involves plotting points on the coordinate plane to reveal the characteristic shape of the parabola. Understanding the structure of the quadratic equation aids in predicting and sketching the graph.
Here鈥檚 how you graph a quadratic:
  • Determine the vertex and mark it on the graph.
  • Check whether the parabola opens upwards or downwards based on the coefficient \( a \).
  • Choose additional x-values and compute their corresponding y-values to get more points.
  • Draw a smooth curve through the points, ensuring symmetry around the vertex.
Graphing helps in visualizing the parabola, making it easier to understand the behavior and solutions to the quadratic equation.
Maximum and Minimum Points
The maximum and minimum points of a parabola are integral to comprehending the function's range and purpose.
Here鈥檚 how they are determined:
  • **Upward-opening parabolas** have a minimum point at the vertex. This is the lowest value the function can reach.
  • **Downward-opening parabolas** have a maximum point at the vertex. This represents the highest value the function can achieve.
The y-coordinate of the vertex indicates the function's maximum or minimum value, depending on the parabola's direction. These points are essential when solving optimization problems and provide insight into applications such as maximizing area or minimizing cost.

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Most popular questions from this chapter

Sketch the graph of each rational function. Specify the intercepts and the asymptotes. (a) \(f(x)=(x-1)(x+2.75) /[(x+1)(x+3)]\) (b) \(g(x)=(x-1)(x+3.25) /[(x+1)(x+3)]\) [Compare the graphs you obtain in parts (a) and (b). Notice how a relatively small change in one of the constants can radically alter the graph.]

(a) Use a graphing utility to draw a graph of each function. (b) For each \(x\) -intercept, zoom in until you can estimate it accurately to the nearest one-tenth. (c) Use algebra to determine each \(x\) -intercept. If an intercept involves a radical, give that answer as well as a calculator approximation rounded to three decimal places. Check to see that your results are consistent with the graphical estimates obtained in part (b). $$N(t)=t^{7}+8 t^{4}+16 t$$

The functions \(f, g,\) and h are defined as follows: $$ f(x)=2 x-3 \quad g(x)=x^{2}+4 x+1 \quad h(x)=1-2 x^{2} $$ In each exercise, classify the function as linear, quadratic, or neither. $$f \circ g$$

The following table and scatter plot show global coal consumption for the years \(1990-1995\). $$\begin{array}{cc} \hline \text { Year } x & \text { Coal consumption } y \\ \hline x=0 \leftrightarrow 1990 & \text { (billion tons) } \\ \hline 0 & 3.368 \\ 1 & 3.285 \\ 2 & 3.258 \\ 3 & 3.243 \\ 4 & 3.261 \\ 5 & 3.311 \\ \hline \end{array}$$ (GRAPH CAN'T COPY) (a) Use a graphing utility to find a quadratic model for the data. Then use the model to make estimates for global coal consumption in 1989 and 1996 (b) Use the following information to show that, in terms of percentage error, the 1996 estimate is better than the 1989 estimate, but in both cases the percentage error is less than \(2 \% .\) The actual figures for coal consumption in 1989 and 1996 are 3.408 and 3.428 billion tons, respectively. (c) Use the model to project worldwide coal consumption in \(1998 .\) Then show that the percentage error is more than \(9 \%,\) given that the actual 1998 consumption was 3.329 billion tons.

Let \(f(x)=\left(x^{5}+1\right) / x^{2}\) (a) Graph the function \(f\) using a viewing rectangle that extends from -4 to 4 in the \(x\) -direction and from -8 to 8 in the \(y\) -direction. (b) Add the graph of the curve \(y=x^{3}\) to your picture in part (a). Note that as \(|x|\) increases (that is, as \(x\) moves away from the origin), the graph of \(f\) looks more and more like the curve \(y=x^{3} .\) For additional perspective, first change the viewing rectangle so that \(y\) extends from -20 to \(20 .\) (Retain the \(x\) -settings for the moment.) Describe what you see. Next, adjust the viewing rectangle so that \(x\) extends from -10 to 10 and \(y\) extends from -100 to \(100 .\) Summarize your observations. (c) In the text we said that a line is an asymptote for a curve if the distance between the line and the curve approaches zero as we move further and further out along the curve. The work in part (b) illustrates that a curve can behave like an asymptote for another curve. In particular, part (b) illustrates that the distance between the curve \(y=x^{3}\) and the graph of the given function \(f\) approaches zero as we move further and further out along the graph of \(f .\) That is, the curve \(y=x^{3}\) is an "asymptote" for the graph of the given function \(f\). Complete the following two tables for a numerical perspective on this. In the tables, \(d\) denotes the vertical distance between the curve \(y=x^{3}\) and the graph of \(f:\) $$ d=\left|\frac{x^{5}+1}{x^{2}}-x^{3}\right| $$ $$\begin{array}{llllll} \hline x & 5 & 10 & 50 & 100 & 500 \\ \hline d & & & & \\ \hline & & & & \\ \hline x & -5 & -10 & -50 & -100 & -500 \\ \hline d & & & & \\ \hline \end{array}$$ (d) Parts (b) and (c) have provided both a graphical and a numerical perspective. For an algebraic perspective that ties together the previous results, verify the following identity, and then use it to explain why the results in parts (b) and (c) were inevitable: $$ \frac{x^{5}+1}{x^{2}}=x^{3}+\frac{1}{x^{2}} $$
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