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3 years ago

10

You start at (5, -3). you move up 8 units. where do you end?

1 answer:

djyliett [7]3 years ago

7 0

Answer:

(5,5)

Step-by-step explanation:

You draw a graph and put an x where (5, -3) is and then from that point, you go up by 8 units

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Which of the following equations could be th equation to represent the given graph? Make sure you explain your answer thoroughly

Alex17521 [72]

Answer:

Option (C) is correct.

Step-by-step explanation:

The given options of the possible equation for the graph are as follows:

$(A) y=2\left(\frac{3}{2}\right)^x \\\\(B) y=-2\left(\frac{3}{2}\right)^{-x} \\\\(C) y=2\left(\frac{2}{3}\right)^x \\\\(D) y=-2\left(\frac{2}{3}\right)^{-x} \\\\$

The given graph is decreasing and at x=0, y=2.

So, first checking the value of the given options for x=0

$(A) y=2\left(\frac{3}{2}\right)^0=2\times 1= 2 \\\\(B) y=--2\left(\frac{3}{2}\right)^{-0}= -2\times 1= -2 \; (not\; possible) \\\\(C) y=2\left(\frac{2}{3}\right)^0= 2\times 1= 2 \\\\(D) y=2\left(\frac{2}{3}\right)^{-0} = -2\times 1= -2 \; (not\; possible)$

As, for x=0, y=2, so options (C) and (D) are not possible, so rejected.

Now, checking the nature (increasing or decreasing) of the given equation by differentiating it.

For option (A),

$\frac{dy}{dx}=2\left(\frac{3}{2}\right)^{x}\times \ln\left(\frac{3}{2}\right)$

As $\ln \left(\frac{3}{2}\right)=\ln(1.5)>0 \;and\; \left(\frac{3}{2}\right)^{x} >0$

So, $\frac{dy}{dx}>0$

Therefore, the function in option (A) is increasing function.

Similarly, for option (C),

$\frac{dy}{dx}=2\left(\frac{2}{3}\right)^{x}\times \ln\left(\frac{2}{3}\right)$

As $\ln \left(\frac{2}{3}\right)=\ln(0.67)0$

So, $\frac{dy}{dx}$

Therefore, the function in option (C) is decreasing function.

As the given graph is decreasing, so, (C) represents $y=2\left(\frac{2}{3}\right)^x$ the given graph.

Hence, option (C) is correct.

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3 years ago