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Chapter 11: Problem 39

Graph each parabola. Give the vertex, axis of symmetry, domain, and range. $$ x=3 y^{2}+12 y+5 $$

Short Answer

Expert verified
Vertex: (2, -3). Axis of Symmetry: \(x = 2\). Domain: \(\{x \in \mathbb{R}\}\). Range: \(\{y \geq -3\}\).

Step by step solution

01

Identify the Vertex

The given equation is in vertex form, which is \[ f(x) = a(x-h)^2 + k \] for a parabola. Here, \(a = \frac{1}{2} \), \(h = 2\), and \(k = -3\). So, the vertex is at the point (h, k) = (2, -3).
02

Determine the Axis of Symmetry

The axis of symmetry for a parabola in vertex form \(f(x) = a(x-h)^2 + k\) is always the vertical line \(x = h\). Therefore, the axis of symmetry is \(x = 2\).
03

Determine the Domain

The domain of any parabola is all real numbers because a parabola extends infinitely in both the left and right directions. So, the domain is \(\{x \in \mathbb{R}\}\).
04

Determine the Range

Since \(a > 0\) (\(a = \frac{1}{2}\) in this case), the parabola opens upwards. The lowest point on this parabola is the vertex \( (2, -3)\). Therefore, the range is all real numbers greater than or equal to -3, represented as \(\{y | y \geq -3\}\).
05

Sketch the Graph

Using the vertex \((2, -3)\), plot the point on a coordinate plane. Draw the axis of symmetry, which is the vertical line \(x = 2\). Since the parabola opens upwards and is wider than the standard \(y = x^2\) parabola (because \(a = \frac{1}{2} < 1\)), draw the parabola opening up from the vertex point.

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Key Concepts

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

vertex form
When graphing a parabola, one of the easiest forms to work with is the vertex form. This form is written as \( f(x) = a(x-h)^2 + k \). Here, \( (h, k) \) represents the vertex of the parabola, and \( a \) determines the width and direction of the parabola's opening.
In our example equation \( f(x) = \frac{1}{2}(x-2)^2 -3 \):
  • \( h = 2 \)
  • \( k = -3 \)
So, the vertex is at the point \( (2, -3) \). This vertex is the minimum point of the parabola since \( a \) is positive, meaning the parabola opens upwards.
Working with the vertex form allows you to directly read the vertex, making it simpler to start your graph.
axis of symmetry
The axis of symmetry is a vertical line that passes through the vertex of the parabola. It divides the parabola into two mirror-image halves. For parabolas written in vertex form \( f(x) = a(x-h)^2 + k \), the axis of symmetry is always the line \( x = h \).
In our example:
Our equation is \( f(x) = \frac{1}{2}(x-2)^2 -3 \), so:
  • \( h = 2 \)
The axis of symmetry is therefore \( x = 2 \). This means that if you draw a vertical line through \( x = 2 \), the parabola will be symmetric around this line.
Knowing the axis of symmetry helps in plotting the parabola accurately.
domain and range
Understanding the domain and range of a parabola is key to knowing where the graph extends. The domain of a parabola covers all possible x-values it can take. Since parabolas stretch infinitely left and right, the domain is always all real numbers, or \( \{x \, | \, x \in \, \mathbb{R}\} \).
For the range, it depends on whether the parabola opens upwards or downwards. This is determined by the parameter \( a \):
  • If \( a > 0 \), the parabola opens upwards
  • If \( a < 0 \), it opens downwards
Since \( a = \frac{1}{2} \) in our equation, the parabola opens upwards. The lowest point is the vertex \( (2, -3) \). Hence, the range is all y-values greater than or equal to -3:
\( \{y \, | \, y \geq -3\} \).
parabola graph
Graphing a parabola involves several steps to ensure accuracy. We'll use our example equation \( f(x) = \frac{1}{2}(x-2)^2 -3 \):
  • Start with plotting the vertex at \( (2, -3) \).
  • Draw the axis of symmetry, a vertical line at \( x = 2 \).
  • Since the parabola opens upwards and \( a = \frac{1}{2} \), it is wider than the standard parabola.
  • Choose additional points to plot, like those to the left and right of the vertex, and mirror them across the axis of symmetry.
  • Draw the curve through these points.
Use these steps to ensure your parabola is plotted correctly. This helps visualize the solutions and the effects of the quadratic parameters on the shape of the graph.

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Most popular questions from this chapter

William Froude was a 19th century naval architect who used the following expression, known as the Froude number, in shipbuilding. $$ \frac{v^{2}}{g \ell} $$ This expression was also used by R. McNeill Alexander in his research on dinosaurs. (Data from "How Dinosaurs Ran," Scientific American.) Use this expression to find the value of \(v\) (in meters per second), given \(g=9.8 \mathrm{~m}\) per sec \(^{2}\) (Round to the nearest tenth.) Triceratops: \(\ell=2.8\) Froude number \(=0.16\)

Solve using the square root property. Simplify all radicals. $$ (x-3)^{2}=25 $$

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ (2 x+1)^{2}=x+4 $$

Find the discriminant. Use it to determine whether the solutions for each equation are A. two rational numbers B. one rational number C. two irrational numbers D. two nonreal complex numbers. Tell whether the equation can be solved using the zero-factor property, or if the quadratic formula should be used instead. Do not actually solve. $$ 18 x^{2}+60 x+82=0 $$

Use the quadratic formula to solve each equation. (All solutions for these equations are real numbers.) $$ x^{2}-4=2 x $$
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