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Chapter 13: Problem 96

Explain how to use \(x=y^{2}+8 y+9\) to find the parabola's vertex.

Short Answer

Expert verified
The vertex of the parabola described by \(x=y^{2}+8 y+9\) is \(-4\) and \(-7\).

Step by step solution

01

Identify the Given Equation

The given equation is \(x=y^{2}+8 y+9\). Since this does not look like the vertex form, it needs to be rearranged.
02

Rewrite the Given Equation

To make it more recognizable, it is advisable to switch the x and y variables. The equation will then be \(y=x^{2}+8 x+9\). Now, it resembles the form of a quadratic equation.
03

Convert to Vertex Form by Completing the Square

To find the vertex, one would need to transform the equation into the vertex form \(y=a(x-h)^{2}+k\). In our case, \(a=1\), and \(h\) and \(k\) need to be calculated by completing the square. First, the constant term 9 is moved to the other side of the equation, resulting in the equation \(y-9=x^{2}+8x\). Then add the square of half the coefficient of x (which is 4) to both sides. The equation now looks like this: \(y-9+16=x^{2}+8x+16\). This simplifies to \(y+7=(x+4)^{2}\).
04

Identify the Vertex

The vertex of the parabola can now be read off directly from the equation as \(-4\) and \(-7\).

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Key Concepts

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

Completing the Square
Completing the square is a method used in algebra to transform a quadratic equation into its vertex form, making it easier to identify the vertex of a parabola. The core idea is to transform the expression on one side of the equation into a perfect square trinomial. Here鈥檚 how you do it:
  • First, ensure the quadratic term has a coefficient of 1. If not, divide the entire equation by the coefficient.
  • Next, take half of the coefficient of the linear term, square it, and add it to both sides of the equation.
  • This creates a perfect square trinomial on one side, allowing it to be rewritten as a squared binomial.
In the given problem, after rearranging the equation to move constant terms, we took half of 8 (which is 4), squared it to get 16, and added it to both sides. This step made the right-hand side of the equation a complete square: \[ (x+4)^2 = x^2 + 8x + 16 \]This method not only helps in finding the vertex but also makes graphing quadratics easier.
Quadratic Equation
A quadratic equation is a second-degree polynomial, typically expressed in the form \( ax^2 + bx + c = 0 \). The highest power of the variable is 2, which is what gives it the name 'quadratic.'Quadratic equations can arise in various mathematical and real-world contexts. Three primary methods are typically used to solve these equations:
  • Factoring: Writing the quadratic as a product of its factors to find the roots.
  • Quadratic Formula: Using \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the solutions for \( x \).
  • Completing the Square: Rewriting the equation so that one side is a perfect square trinomial.
In our exercise, the equation was initially \( y = x^2 + 8x + 9 \). Recognizing this form is crucial as it sets the stage for completing the square or using other methods.
Vertex Form
The vertex form of a quadratic equation is an alternative way of expressing quadratics, particularly useful for graphing. It is expressed as \( y = a(x-h)^2 + k \).In this form:
  • \( (h,k) \) represents the vertex of the parabola. The vertex is a significant point as it is either the highest or the lowest point on the graph, depending on the parabola鈥檚 orientation.
  • \( a \) determines the width and direction of the parabola. If \( a > 0 \), the parabola opens upwards; if \( a < 0 \), it opens downwards.
By converting the given equation \( y = x^2 + 8x + 9 \) into vertex form using the completing the square technique, the equation becomes \( y + 7 = (x + 4)^2 \). Hence, the vertex here is \[ (-4, -7) \].Understanding and utilizing the vertex form is beneficial not only for identifying the vertex but also for understanding the graph's structure.

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Most popular questions from this chapter

Use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? $$y=-x^{2}+4 x-3$$

What is a parabola?

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. $$3 x^{2}=27-3 y^{2}$$

Indicate whether the graph of each equation is a circle, an ellipse, a hyperbola, or a parabola. Then graph the conic section. $$(x-1)^{2}+(y+2)^{2}=16$$

If \(f(x)=\frac{1}{3} x-5,\) find \(f^{-1}(x) .\) (Section 8.4, Example 4)
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