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What is the approximate value of the function at x = -3?

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How high is the line (measured using the y-axis) over the horizontal point (x=) -3?

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Other answers:

He's saying look above x=-3.
well up 1 from -3?
It's slightly higher than 1.
Somewhere around there! It looks closest to the y=1 line. To me it looks like 1.2 or somewhere around 1.2, but your teacher may want you to stick to the drawn lines for accuracy. I don't kno! But it is definately about 1. There's no way to konw exactly what it is without knowing how each horizontal value (such as -3) determines the vertical value (like the one we are trying to estimate).
CliffSedge made a good point, saying that it's greater than 1. It's true to say that this mystery value is between 1 and 2.
so is there a definite answer i can give?
theEric's guess of 1.2 is pretty good. I really doubt it's higher than 1.3 or 1.4.
The whole point of an approximation is that you DON'T give a definite answer. Guess.
Has your teacher gave you any guidelines for estimating? and thank you!
nope :/
Really, I think you're over complicating. It looks sort of like it could be in the vicinity of 1.2 ish. Good enough.
thanks guys!
There is a definite (exact) answer to give if you had the equation. You can make a good guess and be definite about how uncertain you are.
Sometimes teachers will want you to estimate a certain way, so that there is only one answer, and that might be the case here. So, in that sense there may be a definate answer where there is no definate value. I'd say just guess about 1.2! That would be what all my teachers would've liked, I think!
It looks like the equation would be \[y=\frac{-1}{(x-3)}+1\] If you want to get it exactly.
One of the important things about estimation is that you don't want to claim more precision than you actually have. In this situation, with that scale, you can't really get any better than to estimate to ±½ or ±¼ at best. You can definitely do better than rounding to the nearest whole number, so go out to the tenths place and you'll be fine.
CliffSedge proposed an equation that seems right! Notably, the asemptote is at x=2, and his formula yields y=2 at x=2 which looks pretty exact in the picture.
Again, CliffSedge made a great point: about accuracy this time!

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