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Chapter 10: Problem 19

Sketch the graph of the given equation. $$ (x+2)^{2}=8(y-1) $$

Short Answer

Expert verified
It's a parabola opening upwards with vertex (-2, 1), focus (-2, 3), and directrix y = -1.

Step by step solution

01

Recognize the equation type

The given equation \((x+2)^2 = 8(y-1)\) is a parabola. It is in the form \((x-h)^2 = 4p(y-k)\), indicating a parabola that opens vertically.
02

Identify the vertex

The vertex form of a vertical parabola is \((x-h)^2 = 4p(y-k)\). Comparing with the given equation \((x+2)^2 = 8(y-1)\), we see that the vertex \((h, k)\) is \((-2, 1)\).
03

Determine the direction of the parabola

Since the term \((x+2)^2\) is isolated and equals \(8(y-1)\), and because the coefficient of \(y-1\) is positive, the parabola opens upwards.
04

Find the focus and directrix

For a parabola in the form \((x-h)^2 = 4p(y-k)\), the focus is located at \((h, k+p)\) and the directrix is at \(y = k-p\). Thus, \(4p = 8\), so \(p = 2\). The focus is \((-2, 3)\) and the directrix is \(y = -1\).
05

Plot the graph

Plot the vertex at \((-2, 1)\), mark the focus at \((-2, 3)\), and draw the directrix line \(y = -1\). Then, sketch the parabola opening upward through the vertex.

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Key Concepts

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

Vertex Form
The vertex form of a parabola makes it easy to identify its shape and position. A parabola in vertex form looks like \((x-h)^2 = 4p(y-k)\) for a vertically oriented parabola. Here,
  • \(h\) and \(k\) give us the coordinates of the vertex.
  • \(p\) indicates the distance from the vertex to the focus or directrix.
In the context of our example, the equation \((x+2)^2 = 8(y-1)\) is already in vertex form. It's helpful because:
  • We can quickly identify the vertex. By comparing, we find \((h, k) = (-2, 1)\).
  • The vertex \((-2, 1)\) is the turning point of the parabola, showing where it changes direction.
This particular setup makes sketching easier since we have clear steps to follow, helping in visualization.
Focus and Directrix
The focus and directrix define the precise curve of a parabola. They are crucial because they determine how the parabola "bends."
  • Focus: This is a point from which distances to all points on the parabola are measured.
  • Directrix: A line from which distances to all points on the parabola are measured. It lies outside the curve parallel to the axis of symmetry.
In a vertically oriented parabola of form \((x-h)^2 = 4p(y-k)\):
  • The focus is located at \((h, k + p)\).
  • The directrix is \(y = k - p\).
For the given equation \((x+2)^2 = 8(y-1)\), we know:
  • The parameter \(p\) is half of 8, which is 2.
  • Thus, the focus is at \((-2, 3)\) and the directrix is at \(y = -1\).
These components help in shaping the parabolic curve, guiding it from the vertex.
Quadratic Equations
Quadratic equations are polynomials that take the form \(ax^2 + bx + c = 0\). They describe the shape of a parabola on a graph when rewritten in equation form.
  • Parabolas can open upward or downward, horizontally or vertically, depending on how the equation is structured.
  • In our vertex form \((x-h)^2 = 4p(y-k)\), it governs a parabola that is vertical. If it was \((y-k)^2 = 4p(x-h)\), the parabola would open horizontally.
Quadratic equations have many practical applications, such as:
  • Calculating projectile trajectories in physics.
  • Finding maximum or minimum points in profit graphs and other optimization problems.
Understanding their graphing characteristic helps in many real-world scenarios where predicting and visualizing behavior is essential.

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