/*! 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 66 Use a graphing utility to graph ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Use a graphing utility to graph the parabolas.Write the given equation as a quadratic equation in \(y\) and use the quadratic formula to solve for \(y .\) Enter each of the equations to produce the complete graph. $$y^{2}+10 y-x+25=0$$

Short Answer

Expert verified
The equation becomes \(y = 5 \pm \sqrt{x}\). Draw the parabolic graph based on these two potential solutions.

Step by step solution

01

Rearrange the equation in form \(y^{2} = f(x)\)

Rearrange the given equation \(y^{2}+10y-x+25=0\) to form \(y^{2} = f(x)\). Here, we find \(f(x) = x-25\). So the equation becomes \(y^{2} = x-25-10y\).
02

Use the quadratic formula to solve for \(y\)

Next we use the quadratic formula to solve for \(y\) in the equation \(y^{2} = x-25-10y\). The quadratic formula is given by \(y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 1\), \(b =-10\), and \(c =-x+25\). This yields, \(y=\frac{-(-10)\pm\sqrt{(-10)^{2}-(4)(1)(-x+25)}}{2(1)}\). This simplifies to \(y=\frac{10\pm\sqrt{100+4x-100}}{2}=\frac{10\pm\sqrt{4x}}{2}\). Thus, the roots are given by \(y=\frac{10\pm2\sqrt{x}}{2} = 5 \pm\sqrt{x}\).
03

Graph the solutions

Plotting both potential solutions, \(y = 5 + \sqrt{x}\) and \(y = 5 - \sqrt{x}\) would produce the complete graph of the original equation. This will be a parabolic shape, opening to the right since we're squaring the x-term. While graphing, pay special care to the fact that the graph only exists for \(x \geq 0\), since \(x\) is under the square root.

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.

Parabolas
A parabola is a U-shaped curve that can open up, down, left, or right depending on its equation. When we discuss parabolas in the context of quadratic equations, we typically deal with equations in the form of
  • vertical parabolas: \[ y = ax^2 + bx + c \]
  • horizontal parabolas: \[ x = ay^2 + by + c \]
The vertex of the parabola is a crucial feature; it's the point where the parabola changes direction. For our equation \[ y^2 + 10y - x + 25 = 0 \] when rewritten, exhibits a horizontal parabola opening to the right. This is because the squared term is in \(y\), not \(x\). Understanding these directions and vertex points can help visualize and interpret the graph easily.
Graphing Utility
A graphing utility, such as a graphing calculator or an online graphing tool, is essential in visualizing complicated equations like parabolas. When using a graphing utility:
  • Input your equation in its simplest form. For our purpose, this means rewriting equations as \[ y = 5 \pm \sqrt{x} \]
  • Ensure all allowed values for \( x \) are considered. In our scenario, because \( \sqrt{x} \) means the square root of \( x \), graph only for \( x \geq 0 \)
The utility will show two potential parabolic graphs: one for each solution of y. This handy tool supports checking work quickly, especially when manually solving such equations can lead to errors or oversight in domain restrictions.
Completing the Square
Completing the square is a method used to solve quadratic equations, and it helps in converting the equation into a format that can be easily graphed or put into the quadratic formula. This involves making the equation reflect a perfect square trinomial. Let's walk through a brief overview:
  • Start from a quadratic equation, for instance, like our \[ y^2 + 10y - x + 25 = 0 \]
  • Focus on the \(y\) terms: \[ y^2 + 10y \]
  • Add and subtract the square of half the linear coefficient to balance the equation. Here, this means adding and subtracting \( (10/2)^2 = 25 \)
This procedure transforms the equation into a "completed square," which allows us to handle it more easily with the quadratic formula. Understanding this technique is critical when a more structured graph or solution is required in algebra.
Quadratic Equation
Quadratic equations are fundamental algebraic expressions of the form \[ ax^2 + bx + c = 0 \].They result in parabolas when graphed. Solutions to these equations can be found through methods like factoring, completing the square, or using the quadratic formula. The quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] solves any quadratic equation, providing solutions, or 'roots.' Applied to our context:
  • Given \[ y^2 = x - 25 - 10y \]
  • \( a = 1 \), \( b = -10 \), and \( c = -x + 25 \)
Plug these into the formula to solve for two possible \(y\) values. These values are critical in preparing the graphs via solutions \( y = 5 + \sqrt{x} \) and \( y = 5 - \sqrt{x} \). Mastery of applying the quadratic formula is vital since it is a straightforward way to handle a wide range of problems in algebra.

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

Write \(\quad 4 x^{2}-6 x y+2 y^{2}-3 x+10 y-6=0\) as a quadratic equation in \(y\) and then use the quadratic formula to express \(y\) in terms of \(x .\) Graph the resulting two equations using a graphing utility in a \([-50,70,10]\) by \([-30,50,10]\) viewing rectangle. What effect does the \(x y\) -term have on the graph of the resulting hyperbola? What problems would you encounter if you attempted to write the given equation in standard form by completing the square?

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the ellipse and give the location of its foci. $$9 x^{2}+16 y^{2}-18 x+64 y-71=0$$

Radio towers \(A\) and \(B, 200\) kilometers apart, are situated along the coast, with \(A\) located due west of \(B\). Simultaneous radio signals are sent from each tower to a ship, with the signal from \(B\) received 500 microseconds before the signal from \(A\). a. Assuming that the radio signals travel 300 meters per microsecond, determine the equation of the hyperbola on which the ship is located. b. If the ship lies due north of tower \(B,\) how far out at sea is it?

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$4 x^{2}-9 y^{2}+8 x-18 y-6=0$$

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$9 y^{2}-4 x^{2}-18 y+24 x-63=0$$

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.