/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 Determine the input that produce... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Determine the input that produces the largest or smallest output (whichever is appropriate). State whether the output is largest or smallest. $$y=2 x^{2}-4 x+11$$

Short Answer

Expert verified
The minimum output is 9 at \( x = 1 \).

Step by step solution

01

Identify the Type of Quadratic Function

Examine the quadratic function given: \( y = 2x^2 - 4x + 11 \). Since the coefficient of \( x^2 \) is positive, this function represents an upward-opening parabola, which means it has a minimum point (vertex).
02

Calculate the Vertex of the Parabola

To find the vertex, which gives the minimum or maximum point of a parabola, use the formula for the x-coordinate of the vertex: \( x = -\frac{b}{2a} \). Here, \( a = 2 \) and \( b = -4 \). Substitute these values to get \( x = -\frac{-4}{2 \times 2} = \frac{4}{4} = 1 \).
03

Calculate the Minimum Output

Substitute the x-coordinate of the vertex back into the original function to find the minimum output: \( y = 2(1)^2 - 4(1) + 11 = 2 - 4 + 11 = 9 \). The minimum value of the function is \( y = 9 \).
04

Conclusion

The function \( y = 2x^2 - 4x + 11 \) reaches its minimum value at \( x = 1 \), where the minimum output is \( y = 9 \). This is due to the parabola opening upwards, indicating that the minimum point is located at the vertex of the parabola.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Quadratic Function
A quadratic function is a type of polynomial function that is distinguished by its degree, which is 2. This means the highest power of the variable is squared. Quadratic functions are commonly written in the form \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants and \( x \) is the variable. The graph of a quadratic function is known as a parabola, a U-shaped curve that can either open upwards or downwards.
Some key features of quadratic functions include:
  • The vertex, which is the highest or lowest point on the graph.
  • The axis of symmetry, a vertical line that passes through the vertex and divides the parabola into two symmetrical halves.
  • The direction in which the parabola opens, determined by the sign of the coefficient \( a \). If \( a \) is positive, the parabola opens upwards; if negative, it opens downwards.
Minimum Value
The minimum value of a quadratic function is the lowest point that the function can reach, which occurs at the vertex of an upward-opening parabola. The vertex can be found using the formula for the x-coordinate: \( x = -\frac{b}{2a} \). For the given function \( y = 2x^2 - 4x + 11 \), the coefficients are \( a = 2 \) and \( b = -4 \).
Substituting these into the formula, we find:
\[ x = -\frac{-4}{2 \times 2} = \frac{4}{4} = 1 \]
Once the x-coordinate of the vertex is determined, it can be substituted back into the original quadratic equation to find the minimum y-value, which is the actual minimum output of the function. For this function at \( x = 1 \):
\[ y = 2(1)^2 - 4(1) + 11 = 9 \]
Thus, the minimum value of the function is 9, achieved when \( x = 1 \).
Upward-Opening Parabola
An upward-opening parabola is a graphical representation of a quadratic function where the coefficient \( a \) is positive. In the quadratic equation \( y = ax^2 + bx + c \), the parabola opens upwards if \( a > 0 \), forming a U-shape. This shape indicates that the vertex of the parabola is at its lowest point.
Important characteristics of an upward-opening parabola include:
  • The vertex represents the minimum value of the function, as there is no value below this on the y-axis.
  • The two arms of the parabola extend infinitely upwards, meaning that the function will continue to grow in magnitude as \( x \) moves further from the vertex.
  • Such parabolas exhibit symmetry, with the vertex lying on the axis of symmetry, permitting the left and right sides to mirror each other.

Understanding these features is key in analyzing and solving quadratic functions, determining whether a function has a minimum or maximum value, and predicting the general behavior of the function's graph.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The point \(P\) is in the first quadrant on the graph of \(y=1-x^{4} .\) From \(P,\) perpendiculars are drawn to the \(x\) - and \(y\) -axes, thus forming a rectangle. (a) Express the area of the rectangle as a function of a single variable. (b) Use a graphing utility to graph the area function. Then, using the zoom feature, estimate to the nearest hundredth the maximum possible area for the rectangle.

The population \(y\) (in thousands) of a colony of bacteria after \(t\) hr is given by $$ y=(6 t+12) /(2 t+1) $$ where \(t \geq 0\) (a) Find the initial population and the long-term population. Which is larger? (b) Use a graphing utility to graph the population function. Is the function increasing or decreasing? Check that your response here is consistent with your answers in part (a).

Sketch the graph of each rational function. Specify the intercepts and the asymptotes. (a) \(y=3 x /[(x-1)(x+3)]\) (b) \(y=3 x^{2} /[(x-1)(x+3)]\)

Show that for \(m\) and \(n\) positive integers with \(m1 \quad\) then \(x^{m}1\). (c) Using part (a), explain why for \(m\) and \(n\) positive odd integers, with \(m1\) and \(-1

This exercise shows that if we have a table generated by a linear function and the \(x\) -values are equally spaced, then the first differences of the \(y\) -values are constant. (a) In the following data table, the three \(x\) -entries are equally spaced. Compute the three entries in the \(f(x)\) row assuming that \(f\) is the linear function given by \(f(x)=m x+b\) \(f(x)=m x+b .\) (Don't worry about the fact that your answers contain all four of the letters \(m, b, a, \text { and } h .)\) \begin{tabular}{llll} \hline\(x\) & \(a\) & \(a+h\) & \(a+2 h\) \\ \(f(x)\) & & & \\ \hline \end{tabular} (b) Compute the first differences for the three quantities that you listed in the \(f(x)\) row in part (a). (The two first differences that you obtain should turn out to be equal, as required.)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.