Problem 32 Use completing the square to wri... [FREE SOLUTION] (2024)

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Chapter 10: Problem 32

Use completing the square to write each equation in the form \(y=a(x-h)^{2}+k.\) Identify the vertex, focus, and directrix. $$y=x^{2}+2 x-5$$

Short Answer

Expert verified

The equation in vertex form is \( y = (x + 1)^2 - 6 \). The vertex is \( (-1, -6) \), the focus is at \( (-1, -5.75) \), and the directrix is \( y = -6.25 \).

Step by step solution

01

Rewrite the Equation

Start by rewriting the given quadratic equation in the standard form. The given equation is:\[y = x^2 + 2x - 5\]

02

Complete the Square

To complete the square, take the coefficient of x, divide by 2, and square it. The coefficient of x is 2. \[\left(\frac{2}{2}\right)^2 = 1\]Add and subtract this square inside the equation:\[y = x^2 + 2x + 1 - 1 - 5\]

03

Simplify

Rewrite the equation combining the perfect square trinomial and the constants:\[y = (x + 1)^2 - 6\]So, the equation in the form \(y = a(x-h)^2 + k \) is:\[y = (x + 1)^2 - 6\]

04

Identify the Vertex

The equation is now in the form \( y = a(x-h)^2 + k \). The vertex \((h, k)\) is:\[ (h, k) = (-1, -6) \]

05

Determine the Focus and Directrix

Since the quadratic is in the form \( y = a(x - h)^2 + k \), it represents a parabola that opens upwards. To find the focus and directrix, use the formula where the distance \( p \) from the vertex to the focus \( (h, k + p) \) and to the directrix \( y = k - p \):Given the form \( y = (x + 1)^2 - 6 \), the coefficient a is 1.Then, the equation for the vertex form of a parabola is given as:\( x^2 = 4py\).Hence, comparing it with \[ y = (x + 1)^2 - 6\]:Here, \( 4p = 1 \) implying \( p = \frac{1}{4} \).So, the focus is at \( ( -1, -6 + \frac{1}{4}) = (-1, -5.75) \)and the directrix is at \( y = -6 - \frac{1}{4} = -6.25\).

Key Concepts

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

Quadratic Equations

Quadratic equations are polynomial equations of degree 2. This means the highest exponent of the variable (usually denoted as x) is 2. The standard form is \( ax^2 + bx + c = 0 \).

In this equation:

  • \(a\) is the coefficient of \(x^2\).
  • \(b\) is the coefficient of \(x\).
  • \(c\) is the constant term.

The solutions to a quadratic equation can be found using various methods like:

  • Factoring
  • Completing the square
  • Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In cases where we need to analyze parabolas, it's often useful to express quadratic equations in the vertex form, which reveals more about the graph's properties.

Parabola Vertex Form

Converting a quadratic equation to its vertex form helps us easily find key features of a parabola, such as its vertex. The vertex form of a quadratic equation is \( y = a(x-h)^2 + k \).

In this form:

  • \((h, k)\) is the vertex of the parabola.
  • \(a\) determines the direction and width of the parabola:
    • If \(a > 0\), the parabola opens upwards.
    • If \(a < 0\), the parabola opens downwards.

To convert a standard form quadratic to vertex form, we use the method of completing the square. Let's take an example:

  • Given: \(y = x^2 + 2x - 5\)
  • Rewriting it by completing the square:
  • \(y = (x + 1)^2 - 6\)

This vertex form shows that the parabola has a vertex at \((-1, -6)\).

Focus and Directrix

A parabola can also be described in terms of its focus and directrix.

The focus of a parabola is a point from which distances to points on the parabola are measured. The directrix is a line such that the distance of any point on the parabola to the directrix is equal to its distance to the focus. For a parabola in the vertex form \( y = a(x-h)^2 + k \):

  • \( h \) and \( k \) give us the vertex.
  • The distance \( p \) from the vertex to the focus is found using \( 4p = 1/a \).

In our example \( y = (x + 1)^2 - 6 \), \( a \) is 1 which means:

  • \( 4p = 1 \), so \( p = 1/4 \).
  • Focus is at \( ( -1, -6 + \frac{1}{4}) = (-1, -5.75) \).
  • Directrix is at \( y = -6 - \frac{1}{4} = -6.25 \).

Understanding how these elements work together gives us deeper insight into the behavior and properties of parabolas.

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Problem 32 Use completing the square to wri... [FREE SOLUTION] (3)

Most popular questions from this chapter

Write the equation for each circle described. Center \((0,0)\) and passing through \((-3,-4)\)Use the discriminant to identify the type of conic without rotating the axes. $$x^{2}+2 \sqrt{2} x y+y^{2}+1=0$$Find the center and radius of each circle. $$x^{2}+4 x+y^{2}=5$$Find the vertex, axis of symmetry, \(x\) -intercepts, \(y\) -intercept, focus, anddirectrix for each parabola. Sketch the graph, showing the focus anddirectrix. $$y=\frac{1}{2} x^{2}-2$$Write each of the following equations in one of the forms: \(y=a(x-h)^{2}+k, \quad x=a(y-h)^{2}+k\) \(\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1,\) or\((x-h)^{2}+(y-k)^{2}=r^{2}\). Then identify each equation as the equation of a parabola, an ellipse, or acircle. $$2(4-x)^{2}=4-y^{2}$$
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Problem 32 Use completing the square to wri... [FREE SOLUTION] (2024)

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