Problem 27 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The equation describes a parabola opening upwards with vertex at \((-2, 1)\).

Step 1: Rewrite the Equation

Start by rewriting the given equation so that it is easier to identify the type of conic section. The equation is: \( 4x^2 + 16x - 16y + 32 = 0 \).First, divide every term by 4 to simplify:\( x^2 + 4x - 4y + 8 = 0 \).

Step 2: Rearrange into Standard Form

Rearrange the terms to isolate the quadratic in \(x\):\( x^2 + 4x = 4y - 8 \).

Step 3: Complete the Square for x

Complete the square for the \(x\) terms on the left-hand side. Take the coefficient of \(x\) which is 4, halve it to get 2, and square it to get 4.Add and subtract 4 inside the equation:\( (x^2 + 4x + 4) - 4 = 4y - 8 \).This converts to:\( (x+2)^2 - 4 = 4y - 8 \).

Step 4: Simplify and Solve for y

Move \(-4\) to the right side to simplify the equation:\( (x+2)^2 = 4y - 4 \).Add 4 to both sides of 4y - 4 to set it as a perfect square:\( (x+2)^2 = 4(y-1) \).

Step 5: Identify the Conic Section

The equation \( (x+2)^2 = 4(y-1) \) is in the standard form of a parabola, which is \( (x-h)^2 = 4p(y-k) \). The vertex form \( (x - h)^2 = 4p(y-k) \) indicates it is a parabola opening upwards. Here, the vertex is \((-2, 1)\) and \(p\) is 1.

Step 6: Sketch the Graph

Draw the vertex at \((-2, 1)\). Since \(p = 1\), the focus is at 1 unit above the vertex, at \((-2, 2)\), and the directrix is at \(y = 0\). Sketch a parabola opening upwards with the vertex at \((-2, 1)\).

Key Concepts

Graph Sketching

Sketching the graph of an equation is a powerful way to visually understand the mathematical relationship it represents. When sketching a parabola, it is essential to identify key features that guide you:

• The vertex: This is the highest or lowest point of the parabola, depending on its orientation.
• The axis of symmetry: A vertical line that divides the parabola into two mirror-image halves.
• The direction of opening: Indicates whether the parabola opens upwards or downwards.
• The focus and directrix: They define the parabola's shape and orientation.

In our equation, we completed the square to determine that the parabola opens upwards. The vertex at (-2, 1) helps in setting the graph's central point, while the focus at (-2, 2) and directrix at y = 0 further shape the parabola's path. With these details, you can draw a smooth curve, ensuring it passes through the vertex and respects the symmetry around the axis (x = -2). This visual representation solidifies understanding by allowing you to see how these components fit together.

Conic Sections

Conic sections are shapes created by intersecting a plane with a cone. They include parabolas, circles, ellipses, and hyperbolas. Each conic section has a unique equation that defines its set of points:

• A circle equation involves squared terms for both x and y, resulting in a round shape.
• An ellipse, like a stretched circle, has an equation with squared terms for both x and y but with different coefficients.
• A hyperbola has two branches, marked by a minus sign between squared terms of x and y.
• A parabola, our focus, has just one squared term, indicating its U-shape.

For a parabola, the standard form is \((x-h)^2 = 4p(y-k)\) for vertical orientation. The great thing about the standard form is that it provides clear information about the parabola's vertex at (h, k) and its direction of opening, which hinges on p. Thus, transforming our original equation into its standard form helps us to effortlessly classify it among the conic sections as a parabola.

Completing the Square

Completing the square is a fundamental algebraic technique used to simplify quadratic expressions. It is particularly valuable when rewriting equations to better understand the shape of their graphs or solve them. This method involves three main steps:

• Identify the coefficient of the linear term \(x\) in the quadratic expression \(x^2 + bx\).
• Halve this coefficient and square the result.
• Add and subtract this squared value within the expression to form a perfect square trinomial.

In our exercise, for the expression \(x^2 + 4x\), we took the coefficient 4, halved it to get 2, then squared it, equating to 4. By adding and subtracting this 4, we turned \(x^2 + 4x\) into \((x+2)^2\). By completing the square, we expressed the equation in a form that instantly reveals the vertex, transforming complex algebra into a more accessible, geometrically intuitive concept.