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Chapter 13: Problem 56

Give a step-by-step description of how you would go about graphing the parabola \(x^{2}-2 x-4 y-7=0\).

Short Answer

Expert verified
Rewrite, complete the square, identify vertex, plot points, sketch curve.

Step by step solution

01

Rewrite Equation in Standard Form

The given equation is \(x^2 - 2x - 4y - 7 = 0\). To graph a parabola, we need to rewrite this in standard form for a parabola. The equation can be rewritten by organizing and isolating the \(y\) terms: \(x^2 - 2x - 7 = 4y\). Solving for \(y\), we get \(y = \frac{1}{4}(x^2 - 2x - 7)\). This equation is now in the form \(y = a(x - h)^2 + k\).
02

Complete the Square

To further simplify \(y = \frac{1}{4}(x^2 - 2x - 7)\), we will complete the square for the \(x\)-terms. The expression \(x^2 - 2x\) can be transformed by adding and subtracting \(1\) (the square of half the coefficient of \(x\)): \((x - 1)^2 - 1\). Substituting back, we have: \(y = \frac{1}{4}((x - 1)^2 - 1 - 7)\). This simplifies to \(y = \frac{1}{4}(x - 1)^2 - 2\).
03

Identify the Vertex and Direction of Opening

The standard form \(y = \frac{1}{4}(x - 1)^2 - 2\) indicates a parabola with vertex \((h, k) = (1, -2)\) and opens upwards since \(a = \frac{1}{4} > 0\). The coefficient \(\frac{1}{4}\) also determines the width of the parabola.
04

Plot the Vertex

Start by plotting the vertex on the graph at the point \((1, -2)\).
05

Determine and Plot Additional Points

To plot additional points, choose \(x\) values around the vertex and calculate their corresponding \(y\) values from the equation \(y = \frac{1}{4}(x - 1)^2 - 2\). For instance, if \(x = 0\), \(y = \frac{1}{4}(0 - 1)^2 - 2 = -\frac{9}{4}\), plot the point \((0, -\frac{9}{4})\). Do likewise for other values like \(x = 2\) to find \(y = -\frac{9}{4}\) and plot \((2, -\frac{9}{4})\).
06

Sketch the Parabola

Using the vertex and the additional points, draw a smooth curve to depict the parabola. Ensure the curve opens upwards and is symmetric about the vertical line through the vertex \(x = 1\).

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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 helpful technique when working with quadratic expressions, especially when dealing with parabolas. This method makes it easier to convert a quadratic equation into vertex form.
  • To complete the square, consider a quadratic expression of the form: \(x^2 + bx\).
  • The goal is to transform this expression into a perfect square trinomial.
  • You achieve this by adding and subtracting \((\frac{b}{2})^2\) within the expression.
This means if you start with: \(x^2 - 2x\), you calculate \((\frac{-2}{2})^2 = 1\). Hence, you rewrite \(x^2 - 2x\) as \((x-1)^2 - 1\).
By completing the square, you can then express your quadratic in a more convenient form for graphing, called the vertex form.
Standard Form of a Parabola
The standard form of a parabola is crucial for understanding the shape and position of the curve in the coordinate plane. A parabola can be expressed in the form:
  • \(y = a(x-h)^2 + k\)
where \((h, k)\) is the vertex of the parabola, and \(a\) affects the direction and width of the opening. This form reveals several important features:
  • The parabola opens upwards if \(a > 0\) and downwards if \(a < 0\).
  • The vertex form makes it easy to graph since you can immediately see the vertex \((h,k)\).
  • The value of \(a\) also influences the steepness or flatness of the parabola.
For example, the equation \(y = \frac{1}{4}(x-1)^2 - 2\) demonstrates this structure, showing a vertex at \((1, -2)\) and an opening upwards.
Vertex of a Parabola
The vertex of a parabola is a pivotal point that defines the curve's highest or lowest location. It's an easy-to-spot feature on a graph that indicates the maximum or minimum value of the quadratic function.
  • In vertex form \(y = a(x-h)^2 + k\), the vertex is \((h, k)\).
  • From this point, the parabola extends symmetrically.
  • The vertex is also where the axis of symmetry intersects the parabola.
For the equation \(y = \frac{1}{4}(x-1)^2 - 2\), the vertex \((1, -2)\) is easy to locate. Placed at \(x = 1\), it's the point where the parabola makes its sharpest turn. Understanding this concept helps in quickly sketching or analyzing parabolas.
Plotting Points
Plotting points is a straightforward method to visualize the shape of the parabola. Once you have expressed the quadratic equation in a recognizable form, use this simple step to create an accurate graph.
  • Begin by plotting the vertex, your key reference point.
  • Choose several \(x\) values around this point.
  • Calculate the corresponding \(y\) values using the equation.
  • Plot these pairs of \((x, y)\) values.
For example, from the equation \(y = \frac{1}{4}(x-1)^2 - 2\), take \(x = 0\) and find that \(y = -\frac{9}{4}\). Plot \((0, -\frac{9}{4})\) and do similarly for other \(x\) values like \(2\), confirming the symmetry of the parabola around the vertex. Connect these points smoothly to display the complete parabola.

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Most popular questions from this chapter

For Problems 1-14, write the equation of each of the circles that satisfies the stated conditions. In some cases there may be more than one circle that satisfies the conditions. Express the final equations in the form \(x^{2}+y^{2}+D x+E y+F=0\). Tangent to the \(y\) axis, \(x\) intercepts of 2 and 6 See below

Find the vertices, the endpoints of the minor axis, and the foci of the given ellipse, and sketch its graph. See answer section. $$ 7 x^{2}+11 y^{2}=77 $$

Find the vertices, the endpoints of the minor axis, and the foci of the given ellipse, and sketch its graph. See answer section. $$ 4 x^{2}+3 y^{2}=36 $$

Find an equation of the ellipse that satisfies the oiven conditions. Vertices \((\pm 4,0)\), foci \((\pm 2,0) \quad 3 x^{2}+4 y^{2}=48\)

For Problems \(1-30\), find the vertex, focus, and directrix of the given parabola and sketch its graph. $$ x^{2}+4 y-4=0 $$
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