College Algebra: Concepts Through Functions (4th Edition)
College Algebra: Concepts Through Functions (4th Edition)
4th Edition
ISBN: 9780134686967
Author: Michael Sullivan, Michael Sullivan III
Publisher: PEARSON
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Chapter 5, Problem 1RE
To determine

The vertex, focus, and the directrix of the parabola y2=16x and sketch the graph.

Expert Solution & Answer
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Answer to Problem 1RE

The vertex, focus, and the directrix of the parabola y2=16x are V(0,0),F(4,0) and x=4_ respectively.

Explanation of Solution

Formula used:

Equation of a parabola: Vertex at (0,0); Focus on an axis a>0.

VertexFocusDirectrixEquationAxis of symmetryOpens
(0,0)(a,0)x=ay2=4axx-axisRight
(0,0)(a,0)x=ay2=4axx-axisLeft
(0,0)(0,a)y=ax2=4ayy-axisUp
(0,0)(0,a)y=ax2=4ayy-axisDown

Calculation:

The equation of the parabola is y2=16x.

Notice that the equation is of the form, y2=4ax whose vertex is V(0,0), the focus is (a,0) and the directrix is x=a.

Comparing the equation, y2=4ax and y2=16x, 4a=16.

Thus, a=4.

Therefore, the vertex is V(0,0)_, focus is (4,0)_ and the directrix is x=4_. Hence, the parabola opens upward.

.

To draw the graph of the parabola, find two points to the left and right of the parabola.

Substitute, x=4 in y2=16x,

y2=16(4)y2=64y=64y=±8

The points to the left and the right of the focus are, (4,8) and (4,8).

Use the above information and draw the graph of the parabola y2=16x as shown below in Figure 1.

College Algebra: Concepts Through Functions (4th Edition), Chapter 5, Problem 1RE

From Figure 1 note that the parabola opens left.

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Chapter 5 Solutions

College Algebra: Concepts Through Functions (4th Edition)

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