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tresset_1 [31]
3 years ago
12

4. Which is the graph of the function y = −3(2)^x?

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
3 0

Answer:

The answer is letter B

Step-by-step explanation:

Exponential functions have a horizontal asymptote. The equation of the horizontal asymptote is  y =  0 .

Horizontal Asymptote:  

y =  0

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Find the equation of the line<br> Perpendicular to y=2/3x+2 and passes through the point (3,4)
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keeping in mind that perpendicular lines have negative reciprocal slopes, let's check firstly what's the slope of the equation above hmmmm

\bf y = \stackrel{\stackrel{m}{\downarrow }}{\cfrac{2}{3}}x+2\qquad \impliedby \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill

\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{2}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{3}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{3}{2}}}

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\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{4})~\hspace{10em} \stackrel{slope}{m}\implies -\cfrac{3}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{4}=\stackrel{m}{-\cfrac{3}{2}}(x-\stackrel{x_1}{3}) \implies y-4=-\cfrac{3}{2}x-\cfrac{9}{2} \\\\\\ y=-\cfrac{3}{2}x-\cfrac{9}{2}+4\implies y = -\cfrac{3}{2}x-\cfrac{1}{2}

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