Question

sketch the graph of the hyperbola. determine the foci and the equations of the asymptotes.\\(\frac{(x - 3)^2}{16} - \frac{(y - 3)^2}{9} = 1\\)\
choose the correct graph.\
\\(\bigcirc\\) a.\
\\(\bigcirc\\) b.\
\\(\bigcirc\\) c.\
\\(\bigcirc\\) d.\
the foci are \\(\square\\).\
(use a comma to separate answers. type an ordered pair. type an exact answer.)\
the asymptotes are \\(y = \square\\) and \\(y = \square\\).\
(write in slope - intercept form using integers or fractions. if one of the asymptotes has a negative slope, write the equations right side in the second box.)

Explanation:

Step1: Identify Hyperbola Type

The equation is \(\frac{(x - 3)^2}{16}-\frac{(y - 3)^2}{9}=1\), which is a horizontal hyperbola (since \(x\)-term is positive) with center \((h,k)=(3,3)\), \(a^2 = 16\) (so \(a = 4\)), \(b^2=9\) (so \(b = 3\)).

Step2: Find \(c\) (for Foci)

Use \(c^2=a^2 + b^2\). So \(c^2=16 + 9=25\), thus \(c = 5\). Foci are at \((h\pm c,k)=(3\pm5,3)\), which are \((8,3)\) and \((-2,3)\).

Step3: Find Asymptotes

For horizontal hyperbola, asymptotes are \(y - k=\pm\frac{b}{a}(x - h)\). Substituting \(h = 3\), \(k = 3\), \(a = 4\), \(b = 3\), we get \(y - 3=\pm\frac{3}{4}(x - 3)\). Converting to slope - intercept:

• For \(+\frac{3}{4}\): \(y-3=\frac{3}{4}(x - 3)\Rightarrow y=\frac{3}{4}x-\frac{9}{4}+3=\frac{3}{4}x+\frac{3}{4}\)
• For \(-\frac{3}{4}\): \(y - 3=-\frac{3}{4}(x - 3)\Rightarrow y=-\frac{3}{4}x+\frac{9}{4}+3=-\frac{3}{4}x+\frac{21}{4}\) (Wait, correction: \(y-3=-\frac{3}{4}(x - 3)\Rightarrow y=-\frac{3}{4}x+\frac{9}{4}+3=-\frac{3}{4}x+\frac{9 + 12}{4}=-\frac{3}{4}x+\frac{21}{4}\)? No, wait, \(3=\frac{12}{4}\), so \(\frac{9}{4}+\frac{12}{4}=\frac{21}{4}\)? Wait, no, center is \((3,3)\), so when \(x = 3\), \(y = 3\). Let's re - do: \(y-3=\frac{3}{4}(x - 3)\Rightarrow y=\frac{3}{4}x-\frac{9}{4}+3=\frac{3}{4}x+\frac{3}{4}\) (since \(3=\frac{12}{4}\), \(-\frac{9}{4}+\frac{12}{4}=\frac{3}{4}\)). And \(y - 3=-\frac{3}{4}(x - 3)\Rightarrow y=-\frac{3}{4}x+\frac{9}{4}+3=-\frac{3}{4}x+\frac{9 + 12}{4}=-\frac{3}{4}x+\frac{21}{4}\)? Wait, no, that's a mistake. Wait, \(y-3=-\frac{3}{4}(x - 3)\) expands to \(y=-\frac{3}{4}x+\frac{9}{4}+3\). \(3=\frac{12}{4}\), so \(\frac{9}{4}+\frac{12}{4}=\frac{21}{4}\)? But let's check the center: when \(x = 3\), \(y=-\frac{3}{4}(3)+ \frac{21}{4}=-\frac{9}{4}+\frac{21}{4}=\frac{12}{4}=3\), which is correct. Alternatively, maybe I messed up the sign. Wait, the standard form for horizontal hyperbola asymptotes is \(y=k\pm\frac{b}{a}(x - h)\), so \(y = 3\pm\frac{3}{4}(x - 3)\). So \(y=\frac{3}{4}(x - 3)+3=\frac{3}{4}x-\frac{9}{4}+3=\frac{3}{4}x+\frac{3}{4}\) and \(y=-\frac{3}{4}(x - 3)+3=-\frac{3}{4}x+\frac{9}{4}+3=-\frac{3}{4}x+\frac{21}{4}\).

Step4: Identify Correct Graph

A horizontal hyperbola (opens left and right) with center \((3,3)\), \(a = 4\), \(b = 3\). So the graph should open horizontally, center at \((3,3)\). Looking at the options, the correct graph should be the one with horizontal branches, center around \((3,3)\). (Assuming the graphs: A, B, C, D. The one with horizontal opening, center \((3,3)\) – let's assume the correct graph is the one that has the hyperbola opening left and right, centered at \((3,3)\). For example, if option B has the hyperbola opening left and right with center around \((3,3)\), but since we can't see the graphs clearly, but from the equation, it's horizontal, so the correct graph is the one with horizontal branches. But for the foci and asymptotes, we proceed.)

Answer:

The foci are \((-2,3),(8,3)\).
The asymptotes are \(y=\frac{3}{4}x+\frac{3}{4}\) and \(y =-\frac{3}{4}x+\frac{21}{4}\) (or simplified, \(y=\frac{3}{4}(x - 3)+3\) and \(y=-\frac{3}{4}(x - 3)+3\), but in slope - intercept: \(y=\frac{3}{4}x+\frac{3}{4}\) and \(y=-\frac{3}{4}x+\frac{21}{4}\)).

(For the graph, assuming the correct graph is the one with horizontal hyperbola centered at \((3,3)\), opening left and right. If we had to choose from A - D, and based on the equation, the correct graph is the one with horizontal branches, so if, say, option B is the horizontal one, then the answer for the graph is B. But since the graphs are not fully visible, but the foci and asymptotes are as above.)

Foci: \(\boldsymbol{(-2, 3), (8, 3)}\)
Asymptotes: \(\boldsymbol{y=\frac{3}{4}x+\frac{3}{4}}\) and \(\boldsymbol{y=-\frac{3}{4}x+\frac{21}{4}}\) (or \(\boldsymbol{y=\frac{3}{4}(x - 3)+3}\) and \(\boldsymbol{y=-\frac{3}{4}(x - 3)+3}\))

(For the graph, if we assume the correct graph is the one with horizontal opening, center \((3,3)\), so if the options are labeled, and the correct graph is, say, B, then the graph answer is B. But without clear graph details, we focus on foci and asymptotes.)