008A Sample Final A, Question 6

Question: Sketch $4x^{2}+9(y+1)^{2}=36$. Give coordinates of each of the 4 vertices of the graph.

Foundations
1) What type of function is this?
2) What can you say about the orientation of the graph?
Answer:
1) Since both x and y are squared it must be a hyperbola or an ellipse. Since the coefficients of the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y^2} terms are both positive the graph must be an ellipse.
2) Since the coefficient of the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^2} term is smaller, when we divide both sides by 36 the X-axis will be the major axis.

Solution:

Step 1:
We start by dividing both sides by 36. This yields Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{4x^2}{36} + \frac{9(y + 1)^2}{36} = \frac{x^2}{9} + \frac{(y+1)^2}{4} = 1} .

Step 2:
Now that we have the equation that looks like an ellipse, we can read off the center of the ellipse, (0, -1).
From the center mark the two points that are 3 units left, and three units right of the center.
Then mark the two points that are 2 units up, and two units down from the center.

Final Answer:

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