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Q 28.

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Precalculus Enhanced with Graphing Utilities
Found in: Page 643
Precalculus Enhanced with Graphing Utilities

Precalculus Enhanced with Graphing Utilities

Book edition 6th
Author(s) Sullivan
Pages 1200 pages
ISBN 9780321795465

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Short Answer

Find the equation of the parabola described. Find the two points that define the latus rectum, and graph the equation by hand.

Vertex at 0,0, axis of symmetry is the x-axis; containing the point 2,3.

The equation of a parabola is y2=92x. The points are 98, 94 and 98, -94.

The graph of a parabola is :

See the step by step solution

Step by Step Solution

Step 1. Given Information.

The given vertex is at the point 0,0 and the axis of symmetry is the x-axis and containing the point 2,3.

Step 2. Equation of a parabola.

The vertex is at the origin, the axis of symmetry is the x-axis and the graph contains a point in the first quadrant. The general form of the equation is

y2=4ax

Because the point 2,3 is on the parabola, the coordinates x=2,y=3 must satisfy the equation of the parabola. Substitute the values, we get

32=4a29=8aa=98.

The equation will be y2=92x.

Step 3. Latus Rectum.

The focus is at the point 98,0. The two points that determines the latus rectum by letting x=98. Then,

y2=92xy2=9298y2=8116y=±94.

The points are 98, 94 and 98, -94.

Step 4. Graphing Utility.

The graph of a parabola is

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