-18 Points] DETAILS MY NOTES LARPCALC10 10.4.025. ASK YOUR TEACHER Find the center, vertices, foci, and the equations of the asymptotes of the hyperbola. (If an answer does not exist, enter DNE.) $$ \frac{(x-1)^{2}}{4}-\frac{(y+3)^{2}}{1}=1 $$ center $$ (x, y)=(\square) $$ vertices $$ (x, y)=(\square) $$ (smaller $$ x $$-value) $$ (x, y)=(\square) \text { (larger } x \text {-value) } $$ foci $$ (x, y)=(\square) $$ (smaller $$ x $$-value) $$ (x, y)=(\square) \text { (larger } x \text {-value) } $$ asymptotes $$ \square $$ (negative slope) $$ \square $$ (positive slope) Sketch the hyperbola using the asymptotes as an aid.

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Center: $$ (1, -3) $$

Since the hyperbola is of the form  
$$
\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1,
$$
with  
$$
h = 1,\quad k = -3,\quad a^2 = 4\ (a = 2),\quad b^2 = 1\ (b = 1),
$$  
the hyperbola opens horizontally.

**Vertices:**

The vertices are located at $$ (h \pm a, k) $$. Thus,  
$$
\text{Smaller } x\text{-value vertex: } (1-2, -3) = (-1, -3)
$$
$$
\text{Larger } x\text{-value vertex: } (1+2, -3) = (3, -3)
$$

**Foci:**

The foci are at $$ (h \pm c, k) $$, where  
$$
c = \sqrt{a^2+b^2} = \sqrt{4+1} = \sqrt{5}
$$
Thus,  
$$
\text{Smaller } x\text{-value focus: } (1-\sqrt{5}, -3)
$$
$$
\text{Larger } x\text{-value focus: } (1+\sqrt{5}, -3)
$$

**Asymptotes:**

The equations of the asymptotes for a horizontal hyperbola are given by  
$$
y - k = \pm \frac{b}{a}(x-h)
$$
Substitute $$ k = -3 $$, $$ b = 1 $$, $$ a = 2 $$, and $$ h = 1 $$:  
$$
y + 3 = \pm \frac{1}{2}(x-1)
$$
Thus, the asymptotes are:  
$$
y+3 = -\frac{1}{2}(x-1) \quad \text{(negative slope)}
$$
$$
y+3 = \frac{1}{2}(x-1) \quad \text{(positive slope)}
$$
Center: $$ (1, -3) $$ Vertices: - Smaller $$ x $$-value: $$ (-1, -3) $$ - Larger $$ x $$-value: $$ (3, -3) $$ Foci: - Smaller $$ x $$-value: $$ (1-\sqrt{5}, -3) $$ - Larger $$ x $$-value: $$ (1+\sqrt{5}, -3) $$ Asymptotes: - Negative slope: $$ y + 3 = -\frac{1}{2}(x - 1) $$ - Positive slope: $$ y + 3 = \frac{1}{2}(x - 1) $$

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