Problem 28 - Calculus Early Transcendentals: Pearson New International Edition

Question

Sketch the graph of the given equation. \(x^{2}-4 x+8 y=0\)

Step-by-Step Solution

Answer

The equation represents a downward-opening parabola with vertex at \((2, \frac{1}{2})\).

Step 1: Rewrite the equation

Start by rewriting the equation of the form given: \[ x^2 - 4x + 8y = 0 \]. To easily plot the graph, let's express it in terms of y by isolating y on one side: Add and subtract 4 on both sides to complete the square: \[ x^2 - 4x + 4 - 4 + 8y = 0 \]. This simplifies as: \[ (x-2)^2 - 4 + 8y = 0 \] or \[ (x-2)^2 = 4 - 8y \]. Finally, solve for y: Divide everything by -8: \[ y = \frac{4 - (x-2)^2}{8} \]. Now we can analyze this equation before sketching the graph.

Step 2: Identify the vertex and orientation

The equation \[ y = \frac{4 - (x-2)^2}{8} \] represents a parabola. The general form of a parabola \( y = a(x-h)^2 + k \) indicates the vertex at \((h, k)\). Here, \( h = 2 \) and \( k = \frac{4}{8} = \frac{1}{2} \). So, the vertex is \((2, \frac{1}{2})\). The term \(-\frac{1}{8}(x-2)^2\) implies that the parabola opens downwards because the coefficient is negative.

Step 3: Sketch the graph

Plot the vertex of the parabola at \((2, \frac{1}{2})\). Since it opens downwards due to the negative coefficient in front of \((x-2)^2\), sketch the parabola such that it forms a 'U' shape, inverted or upside down. Ensure it is vertically stretched slightly due to the coefficient \(-\frac{1}{8}\) making the 'U' wider. As \( x \) moves away from \( 2 \), 'U' descends further.

Key Concepts

Parabolas

Parabolas are fascinating shapes that often appear in mathematics when dealing with quadratic functions. They have a distinct U-shaped curve, which can either open upwards or downwards. A parabola is defined by its standard equation, typically written as either

• \( y = ax^2 + bx + c \) for vertical parabolas
• or \( x = ay^2 + by + c \) for horizontal ones.

Understanding the properties of a parabola is crucial. They have a vertex, which is a point representing its highest or lowest value. The line of symmetry, which is a vertical line that passes through the vertex, divides the parabola into two mirror-image halves. Additionally, the direction in which a parabola opens is dictated by the sign of the coefficient 'a' in its equation. If 'a' is positive, it opens upwards. If 'a' is negative, it opens downwards.

Vertex form

The vertex form of a parabola provides a clear picture of its properties. It is expressed as:\[ y = a(x-h)^2 + k \]In this equation,

• \( (h, k) \) is the vertex of the parabola, a point where it turns.
• 'a' determines whether the parabola opens upwards or downwards and affects the shape or stretch.

The vertex form is often preferred for graphing because it directly tells you the vertex's location on a graph. This makes it much simpler to place the vertex in its correct spot and judge the shape and orientation of the parabola without converting from another form. Identifying the vertex helps in sketching more accurately, as it is the pivot from which the curve extends.

Completing the square

Completing the square is a technique used to transform a quadratic equation into the vertex form. This involves a few specific steps, easily broken down. Let's consider the original equation \( x^2 - 4x + 8y = 0 \).

• Firstly, focus on the quadratic terms \( x^2 - 4x \).
• To complete the square, take half of the coefficient of x, square it, and add it inside the equation. Here, half of -4 is -2, and its square is 4.
• Add and subtract 4: \( x^2 - 4x + 4 - 4 + 8y = 0 \).
• This manipulation allows us to rewrite the equation as: \( (x-2)^2 - 4 + 8y = 0 \).

The equation can now be reorganized to show its vertex form, providing a straightforward pathway to graphing. Completing the square simplifies understanding the parabola's properties and aid in its sketching.

Graph sketching

Graph sketching involves placing the key components from an equation onto a coordinate system to visualize a function. For parabolas, graph sketching starts by plotting the vertex.

• In the given scenario, the vertex is located at \( (2, \frac{1}{2}) \).
• Once the vertex is set, observe the 'a' value in the vertex form \( y = a(x-h)^2 + k \) to determine the parabola's direction and width.
• Here, the value of \( -\frac{1}{8} \) indicates an upside-down U-shape, due to the negative sign.
• A smaller absolute value of 'a' suggests the parabola is wider than the standard \( y = x^2 \).

Mark this shape carefully, illustrating how it symmetrically extends from the vertex. As the vertex is the focal point, ensure it's accurately placed, and draw the arms of the parabola that either rise or fall depending on the 'a' value. This visual representation helps deepen understanding of parabolas and their behaviors as mathematical functions.