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Chapter 10: Problem 27

Sketch the graph of the given equation. $$ 4 x^{2}+16 x-16 y+32=0 $$

Short Answer

Expert verified
The vertex-form equation is \( y = \frac{1}{4}(x + 2)^2 + 1 \) with vertex at \((-2, 1)\).

Step by step solution

01

Rewrite the Equation

Begin by rewriting the equation given in the problem: \[ 4x^2 + 16x - 16y + 32 = 0 \]Rearrange the terms to form a complete square for the quadratic terms in \(x\). Move the constant and \(y\)-terms to the right side:\[ 4x^2 + 16x = 16y - 32 \]
02

Complete the Square for x

Factor out the 4 from the \(x\)-terms:\[ 4(x^2 + 4x) = 16y - 32 \]Complete the square inside the parentheses by adding and subtracting \(4\) (since \( (4/2)^2 = 4 \)):\[ 4(x^2 + 4x + 4 - 4) = 16y - 32 \]This gives:\[ 4((x+2)^2 - 4) = 16y - 32 \]
03

Simplify the Expression

Distribute the 4 and simplify:\[ 4(x + 2)^2 - 16 = 16y - 32 \]Add 16 to both sides to isolate the perfect square:\[ 4(x + 2)^2 = 16y - 16 \]
04

Solve for y

To express \(y\) in terms of \(x\), divide every term by 16 to solve for \(y\):\[ y = \frac{4(x + 2)^2 + 16}{16} \]\[ y = \frac{1}{4}(x + 2)^2 + 1 \]
05

Identify the Transformation of the Parent Function

The equation \( y = \frac{1}{4}(x + 2)^2 + 1 \) is a transformation of the parent function \( y = x^2 \).- The \(+2\) inside the square shifts the graph 2 units to the left.- The factor \(\frac{1}{4}\) vertically compresses the parabola by a factor of 4.- The \(+1\) moves the graph 1 unit up.The vertex of the parabola is at \((-2, 1)\).
06

Sketch the Graph

Using the transformations, sketch the graph:1. Start with the vertex at \((-2, 1)\).2. The parabola opens upwards since the coefficient of the squared term is positive.3. Due to vertical compression, the parabola is wider than \(y = x^2\).4. The horizontal shift moves the graph left by 2 units, and the upward shift moves it up by 1 unit.

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Key Concepts

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

Parabola
A parabola is a U-shaped curve that can open upwards or downwards. It is the graph of a quadratic function, generally in the form of \( y = ax^2 + bx + c \). Parabolas have some unique properties that make them stand out in mathematics. The most crucial parts to understand are:
  • Vertex: The highest or lowest point on the parabola, depending on whether it opens upwards or downwards. It's given by \( ext{vertex} = (h, k) \).
  • Axis of Symmetry: A vertical line that passes through the vertex, dividing the parabola into two mirror-image halves. For an equation \( ax^2 + bx + c \), it's given by \( x = -\frac{b}{2a} \).
  • Direction: If the \( a \) value is positive, the parabola opens upwards. If negative, it opens downwards.
The graph of a parabola can be quickly identified and drawn if its vertex and direction are known. Understanding these traits is essential when sketching any parabola from a quadratic equation.
Completing the Square
Completing the square is a method used to transform a quadratic equation into a perfect square trinomial. This is helpful for graphing parabolas and solving quadratic equations by making them easier to manipulate.To complete the square for a quadratic expression like \( x^2 + bx \), you.
  • Find half of the \( b \) coefficient and square it: \( (\frac{b}{2})^2 \).
  • Add and subtract this square inside the expression, creating a perfect square trinomial.
  • Factor the trinomial into \( (x + \frac{b}{2})^2 \) and adjust as needed to maintain equality.
For example, with \( x^2 + 4x \), you find \( (4/2)^2 = 4 \), resulting in \( (x+2)^2 - 4 \). Completing the square is a powerful tool for rewriting quadratics in vertex form, revealing important features like the vertex coordinates and simplifying graph drawing.
Quadratic Transformations
Quadratic transformations involve modifying the basic parabola \( y = x^2 \) to shift, stretch, or compress it. These transformations help graph different quadratic functions based on given equations.The main transformations include:
  • Vertical Shifts: Adding or subtracting a constant \( c \) moves the graph up or down, e.g., \( y = x^2 + c \).
  • Horizontal Shifts: Adjustments inside the square, like \( y = (x + h)^2 \), shift the graph left (\(-h\)) or right (\(+h\)).
  • Vertical Stretch and Compression: Multiplying the squared term by a factor \( a \) stretches (\( a > 1 \)) or compresses (\( 0 < a < 1 \)) the parabola.
  • Reflection: A negative \( a \) value reflects the parabola over the x-axis.
An understanding of these transformations allows you to predict and sketch the curve without needing to plot numerous points manually. In the exercise, the graph was shifted and compressed, leading to the equation \( y = \frac{1}{4}(x + 2)^2 + 1 \) as a variation of the parent function.

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Most popular questions from this chapter

In Problems 35-46, find the length of the parametric curve defined over the given interval. $$ x=\tanh t, y=\ln \left(\cosh ^{2} t\right) ;-3 \leq t \leq 3 $$

The path of a projectile fired from level ground with a speed of \(v_{0}\) feet per second at an angle \(\alpha\) with the ground is given by the parametric equations $$ x=\left(v_{0} \cos \alpha\right) t, \quad y=-16 t^{2}+\left(v_{0} \sin \alpha\right) t $$ (a) Show that the path is a parabola. (b) Find the time of flight. (c) Show that the range (horizontal distance traveled) is \(\left(v_{0}^{2} / 32\right) \sin 2 \alpha\). (d) For a given \(v_{0}\), what value of \(\alpha\) gives the largest possible range?

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