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

Problem

((x−1)2)/16+((y+2)2)/9=1

Solution

1. Identify the shape by comparing the equation to the standard form of a conic section. Since both variables are squared, have positive coefficients, and are set equal to 1, this is a horizontal ellipse of the form ((x−h)2)/(a2)+((y−k)2)/(b2)=1

2. Determine the center (h,k) by looking at the terms (x−1) and (y+2) The center is (1,−2)

3. Calculate the semi-axes lengths by taking the square roots of the denominators. The horizontal semi-major axis is a=√(,16)=4 and the vertical semi-minor axis is b=√(,9)=3

4. Find the vertices by moving a units left and right from the center. The vertices are (1±4,−2) which results in (−3,−2) and (5,−2)

5. Find the co-vertices by moving b units up and down from the center. The co-vertices are (1,−2±3) which results in (1,−5) and (1,1)

6. Locate the foci using the relationship c2=a2−b2 Since c2=16−9=7 the distance to the foci is c=√(,7)≈2.65 The foci are located at (1±√(,7),−2)

7. Sketch the graph by plotting the center, vertices, and co-vertices, then drawing a smooth curve to connect the four outer points.

Final Answer

((x−1)2)/16+((y+2)2)/9=1* is an ellipse centered at *(1,−2)* with vertices *(5,−2),(−3,−2)* and co-vertices *(1,1),(1,−5)

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