Question

match the graphs to their equations.
\\( \frac{(x - 2)^2}{16}+\frac{(y + 1)^2}{16}=1 \\)
\\( \frac{(x + 2)^2}{16}+\frac{(y - 1)^2}{16}=1 \\)
\\( \frac{(x - 2)^2}{16}+\frac{(y - 1)^2}{16}=1 \\)
\\( \frac{(x + 2)^2}{16}+\frac{(y + 1)^2}{16}=1 \\)
question help: \\( \square \\) video \\( \square \\) post to forum

Explanation:

Step1: Recall the standard form of a circle equation

The standard form of a circle equation is \(\frac{(x - h)^2}{r^2}+\frac{(y - k)^2}{r^2}=1\), where \((h,k)\) is the center of the circle.

Step2: Analyze the first equation \(\frac{(x - 2)^2}{16}+\frac{(y + 1)^2}{16}=1\)

Here \(h = 2\), \(k=-1\).

Step3: Analyze the second equation \(\frac{(x + 2)^2}{16}+\frac{(y - 1)^2}{16}=1\)

Here \(h=-2\), \(k = 1\).

Step4: Analyze the third equation \(\frac{(x - 2)^2}{16}+\frac{(y - 1)^2}{16}=1\)

Here \(h = 2\), \(k = 1\).

Step5: Analyze the fourth equation \(\frac{(x + 2)^2}{16}+\frac{(y + 1)^2}{16}=1\)

Here \(h=-2\), \(k=-1\).

Step6: Match the centers to the graphs

• For \(h = 2\), \(k=-1\), it matches graph \(b\).
• For \(h=-2\), \(k = 1\), it matches graph \(c\).
• For \(h = 2\), \(k = 1\), it matches graph \(a\).
• For \(h=-2\), \(k=-1\), it matches graph \(d\).

Answer:

\(\frac{(x - 2)^2}{16}+\frac{(y + 1)^2}{16}=1\) matches \(b\); \(\frac{(x + 2)^2}{16}+\frac{(y - 1)^2}{16}=1\) matches \(c\); \(\frac{(x - 2)^2}{16}+\frac{(y - 1)^2}{16}=1\) matches \(a\); \(\frac{(x + 2)^2}{16}+\frac{(y + 1)^2}{16}=1\) matches \(d\)