/*! 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 113 Sketch a graph of the quadratic ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 113

Sketch a graph of the quadratic function and give the vertex, axis of symmetry, and intercepts. \(f(x)=x^{2}-5 x-6\)

Short Answer

Expert verified
Vertex: \((\frac{5}{2}, -\frac{49}{4})\), axis of symmetry: \(x = \frac{5}{2}\), y-intercept: \((0, -6)\), x-intercepts: \((6, 0)\) and \((-1, 0)\).

Step by step solution

01

Determine the Vertex

The vertex of a quadratic function in the form \(f(x) = ax^2 + bx + c\) can be found using the formula \(x = -\frac{b}{2a}\). Here, \(a = 1\), \(b = -5\), and \(c = -6\). Calculating this, we get \(x = -\frac{-5}{2 \times 1} = \frac{5}{2}\). To find the \(y\)-coordinate of the vertex, substitute \(x = \frac{5}{2}\) back into the function: \(f(\frac{5}{2}) = (\frac{5}{2})^2 - 5(\frac{5}{2}) - 6 = \frac{25}{4} - \frac{25}{2} - 6 = -\frac{49}{4}\). Therefore, the vertex is \((\frac{5}{2}, -\frac{49}{4})\).
02

Find the Axis of Symmetry

The axis of symmetry for a parabola described by \(f(x) = ax^2 + bx + c\) is the vertical line \(x = -\frac{b}{2a}\). Thus, from Step 1, the axis of symmetry is \(x = \frac{5}{2}\).
03

Calculate the Y-intercept

The y-intercept occurs where \(x = 0\). Substitute \(x = 0\) into the function: \(f(0) = 0^2 - 5 \cdot 0 - 6 = -6\). Therefore, the y-intercept is \((0, -6)\).
04

Determine the X-intercepts

Find the x-intercepts by solving the equation \(x^2 - 5x - 6 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = -5\), and \(c = -6\). Calculate the discriminant: \(b^2 - 4ac = (-5)^2 - 4 \cdot 1 \cdot (-6) = 25 + 24 = 49\). Thus, \(x = \frac{5 \pm \sqrt{49}}{2} = \frac{5 \pm 7}{2}\). This gives us two solutions: \(x = 6\) and \(x = -1\). Therefore, the x-intercepts are \((6, 0)\) and \((-1, 0)\).
05

Sketch the Graph

To sketch the parabola, plot the vertex \((\frac{5}{2}, -\frac{49}{4})\), the y-intercept \((0, -6)\), and the x-intercepts \((6, 0)\) and \((-1, 0)\) on a coordinate plane. Draw a smooth curve passing through these points to form a parabola opening upwards. The axis of symmetry \(x = \frac{5}{2}\) can also be drawn as a dashed vertical line.

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 91影视!

Key Concepts

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

Vertex of a Parabola
The vertex of a quadratic function is a crucial point that identifies the extremity of the curve; either the highest or lowest point depending on its orientation.In the function given by \(f(x) = x^2 - 5x - 6\), we find the vertex using the formula \(x = -\frac{b}{2a}\).
Here, \(a = 1\) and \(b = -5\), giving us \(x = \frac{5}{2}\).
Substitute this back into the function to find the \(y\)-coordinate:
  • \(f(\frac{5}{2}) = (\frac{5}{2})^2 - 5(\frac{5}{2}) - 6\)
  • This simplifies to \(f(\frac{5}{2}) = -\frac{49}{4}\).
Thus, the vertex is located at \((\frac{5}{2}, -\frac{49}{4})\), a vital point for sketching the graph.
Axis of Symmetry
The axis of symmetry is a vertical line that passes through the vertex of the parabola, neatly bisecting it into two mirror-image halves.
For a quadratic function of the form \(f(x) = ax^2 + bx + c\), the axis of symmetry is given by \(x = -\frac{b}{2a}\).
From our previous calculation, we get \(x = \frac{5}{2}\).
This means the axis of symmetry for our quadratic function is the line \(x = \frac{5}{2}\).
  • It helps guide the symmetry in your graphing as all points on the parabola are symmetrical about this vertical line.
  • In sketches, it can be handy to represent this line with a dashed line for additional clarity.
Y-intercept
The y-intercept of a parabola is the point where it crosses the y-axis, occurring when \(x = 0\).
To find this for \(f(x) = x^2 - 5x - 6\), simply substitute \(x = 0\):
  • \(f(0) = 0^2 - 5 \cdot 0 - 6 = -6\)
Thus, the y-intercept is \((0, -6)\).This point is crucial because it indicates where the graph starts on the y-axis, providing you additional information for plotting.
X-intercepts
X-intercepts are the points where the parabola crosses the x-axis, and they are found by setting the quadratic function \(f(x)\) equal to zero.
For the equation \(x^2 - 5x - 6 = 0\), we use the quadratic formula:
  • \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
With \(a = 1\), \(b = -5\), and \(c = -6\), we calculate:
  • The discriminant: \(b^2 - 4ac = 49\)
  • Solutions: \(x = \frac{5 \pm 7}{2}\) gives us \(x = 6\) and \(x = -1\)
Thus, the x-intercepts are at \((6, 0)\) and \((-1, 0)\).These points reflect the roots of the equation, telling us where the curve intersects the x-axis.
Graphing Quadratic Functions
Graphing quadratic functions involves plotting key points such as the vertex, axis of symmetry, and intercepts to shape the characteristic "U" shape.
Here's how to do it:
  • Plot the vertex \((\frac{5}{2}, -\frac{49}{4})\), a significant point that serves as a reference for the parabola's direction and width.
  • Draw the axis of symmetry \(x = \frac{5}{2}\), helping you ensure the parabola is symmetric.
  • Mark the y-intercept \((0, -6)\) and x-intercepts \((6, 0)\) and \((-1, 0)\) on the coordinate plane.
  • Connect these points with a smooth, curved line opening upwards, flowing through these calculated points.
This approach provides a visual representation of the function and offers insights into the behavior of quadratic equations.

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

For the following exercises, use the given information to answer the questions. The distance \(s\) that an object falls varies directly with the square of the time, \(t,\) of the fall. If an object falls 16 feet in one second, how long for it to fall 144 feet?

Use the given volume and radius of a cylinder to express the height of the cylinder algebraically. Volume is \(\pi\left(3 x^{4}+24 x^{3}+46 x^{2}-16 x-32\right),\) radius is \(x+4.\)

For the following exercises, use the given information to answer the questions. The weight of an object above the surface of the Earth varies inversely with the square of the distance from the center of the Earth. If a body weighs 50 pounds when it is 3960 miles from Earth鈥檚 center, what would it weigh it were 3970 miles from Earth鈥檚 center?

Use synthetic division to determine whether the first expression is a factor of the second. If it is, indicate the factorization. $$x+\frac{1}{3}, 3 x^{4}+x^{3}-3 x+1$$

Use synthetic division to find the quotient $$\left(x^{4}+5 x^{3}-3 x^{2}-13 x+10\right) \div(x+5)$$
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation
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.