Find the vertices of the ellipse define \( \frac{(x+4)^{2}}{4}+\frac{(y-1)^{2}}{25}=1 \). If necessary, ro
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To find the vertices of the given ellipse, we start with the standard equation: \[ \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \] In your equation, \(\frac{(x+4)^2}{4} + \frac{(y-1)^2}{25} = 1\), we can identify \(h = -4\), \(k = 1\), \(a^2 = 4\) (so \(a = 2\)), and \(b^2 = 25\) (so \(b = 5\)). The center of the ellipse is at the point \((-4, 1)\). Since \(b > a\), the ellipse is vertical. The vertices are found by adding and subtracting \(b\) from the \(y\) coordinate of the center: 1. Top vertex: \((-4, 1 + 5) = (-4, 6)\) 2. Bottom vertex: \((-4, 1 - 5) = (-4, -4)\) Now, you have the vertices at the points \((-4, 6)\) and \((-4, -4)\). Enjoy your exploration of ellipses! Always remember to pay attention to the orientation based on the values of \(a\) and \(b\); a simple switch can lead you down a completely different path!