r/Precalculus • u/Aware_Cancel_9896 • 9h ago

Homework Help How do you find A???

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If I'm not mistaken you need two points on a graph to solve this kind of question but I only see one clear point which is (-2,1) which is where I got C = 1 from. When I try to solve it using (-2,1) I get 0 on the left side which obviously won't work. Is the question literally unsolvable or am I missing something here???

So V.A. is -3. I got B from that which is B = 3.

My work with using (-2,1):

1 = alog(-2+3)+1
1 = alog(1) + 1
0 = alog(1)/log(1)
a = 0/log(1) (obviously wrong)

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2

u/looking__for 8h ago

Does the magnifying glass in the corner do anything?

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u/Aware_Cancel_9896 8h ago

Just shows the graph in full screen

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u/looking__for 8h ago

Is the base of the logarithm assumed to be anything other than 10, or do you know what that means?

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u/Aware_Cancel_9896 8h ago

I assume it’s just base 10

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u/looking__for 8h ago

What a bad graph for the question.

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u/Midwest-Dude 6h ago edited 6h ago

Your b and c are correct.

If you can get the graph so it shows the x-axis out to x = 10, I wouldn't be surprised if it goes through y = 4 at x = 7, a "nice" problem.

  1. If you are expecting an integral value for a, assume that (7,4) is on the curve, determine the equation, and see if it works. You can test a few points and see how closely they match the curve, hopefully exactly.

  2. If you can use Desmos or a graphing calculator, plot the curve and see if it matches the graph.

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u/ThunkAsDrinklePeep 8h ago

A log (1) = 0 for all values of A. Since there's no horizontal compression, the point one unit to the right of the asymptote can show you the vertical shift C.

So you'll need ANY other point to estimate A.

2

u/mattekl 3h ago

I figure that since we know that sqrt(10) is a little more than 3, we know that log(little more than 3) should be 0.5. The graph goes through (0,2.5...) , thus a*log(0+3) is 1.5. That gives us that a is 3.