The correct graph of the ellipse is identified and attached below.
Given the equation is (x₋3)²/16 ₊ (y₋4)²/9 = 1
step 1: identify the center (h,k) of the ellipse in standard form:
(x₋h)²/a² ₊ (y₋k)²/b² = 1
In the standard form of an ellipse, notice that there is a minus sign in front of the h and k.
In our equation (x₋3)²/16 ₊ (y₋4)²/9 = 1 , we have:
h = 3 and k = 4
Therefore, center (h,k) = (3,4)
Step 2: Identify the orientation by determining the major axis and minor axis in (x₋h)²/a² ₊ (y₋k)²/b² = 1
if a>b, the major axis is horizontal and the minor axis is vertical, making the ellipse a horizontal ellipse.
if a<b, the minor axis is horizontal and the major axis is vertical, making the ellipse a vertical ellipse.
For (x₋3)²/16 ₊ (y₋4)²/9 = 1
a² = 16, a = 4 and b² = 9 , b = 3
so, 4>3, the major axis is horizontal and the minor axis is vertical, making the ellipse a horizontal ellipse.
Step 3:
Plot the co-vertices along the minor axis and the vertices along the main axis (endpoints). No matter which way around, draw points a horizontal distance of a to the left and right from the center and points b above and below the center.
Therefore the vertices are marked along the major axis.
vertices are (3,7) and (3,1)
and co vertices are marked along the minor axis
co vertices are (₋1,4 ) and (7,4)
step 4: connect the vertices and co vertices making a smooth horizontal ellipse.
Hence a horizontal ellipse is made.
Learn more about Ellipse here:
brainly.com/question/4429071
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