Graph (x^2)/16-(y^2)/9=1

Problem

(x2)/16−(y2)/9=1

Solution

1. Identify the type of conic section. Since the equation is in the form (x2)/(a2)−(y2)/(b2)=1 it represents a horizontal hyperbola centered at the origin (0,0)

2. Determine the values of a and b From the denominators, a2=16 and b2=9 which gives a=4 and b=3

3. Locate the vertices. The vertices are located at (±a,0) which are (4,0) and (−4,0)

4. Find the asymptotes. The equations for the asymptotes of a horizontal hyperbola are y=±b/a*x Substituting the values gives y=±3/4*x

5. Calculate the foci. Use the relationship c2=a2+b2 Here, c2=16+9=25 so c=5 The foci are at (±5,0)

6. Sketch the graph. Draw the central rectangle using x=±4 and y=±3 draw the diagonal asymptotes through the corners, and plot the vertices to draw the two branches of the hyperbola opening left and right.

Final Answer

(x2)/16−(y2)/9=1* is a horizontal hyperbola with vertices *(±4,0)* and asymptotes *y=±3/4*x

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