Thread: Evaluate the function at the given numbers (correct to six decimal places).

Evaluate the function at the given numbers (correct to six decimal places).

f(x) = (x^2 - 6x)/ (x^2 -x -2)


x= -0.9 y: -21.41

Webassign marked this as incorrect. How is this incorrect? I plug that exact value into the the equation on my ti 83 and got that answer. Webassign also marked me wrong for other decimal values except for
0, -.5, and 2

Originally Posted by NeoSonata

Evaluate the function at the given numbers (correct to six decimal places).

f(x) = (x^2 - 6x)/ (x^2 -x -2)


x= -0.9 y: -21.41

Webassign marked this as incorrect. How is this incorrect? I plug that exact value into the the equation on my ti 83 and got that answer. Webassign also marked me wrong for other decimal values except for
0, -.5, and 2

Are those answers to six decimal places?

Originally Posted by Prove It

Are those answers to six decimal places?

ah, I was using the table feature of the ti83 which didn't show the full number.

How dumb of me.

Originally Posted by NeoSonata

ti83 only showed those numbers.

One of my answer was -2.6 which i got by plugging -0.5 in the equation and that got marked correct

Then you must need to show more floating point numbers in your answers. You can change this in the settings.

Okay I changed it in the setting to "float" but some numbers are not reaching the 6 decimal place.

For value x value -0.9 i got -21.41379.

I plug that entire number into the webassign and it still marked me incorrect.
I also tried adding 0 at the end to meet the "6 decimal place" but that failed.

Originally Posted by NeoSonata

Okay I changed it in the setting to "float" but some numbers are not reaching the 6 decimal place.

For value x value -0.9 i got -21.41379.

I plug that entire number into the webassign and it still marked me incorrect.
I also tried adding 0 at the end to meet the "6 decimal place" but that failed.

Because that is to five decimal places, if to six decimal places it is -21.413793.

CB

Originally Posted by CaptainBlack

Because that is to five decimal places, if to six decimal places it is -21.413793.

CB

My ti83 will not show the 6 decimal place when I input it into the table or trace it on the graph. any thoughts?

Originally Posted by NeoSonata

My ti83 will not show the 6 decimal place when I input it into the table or trace it on the graph. any thoughts?

Here's a thought, old fashioned as it is: Type in ((-0.9)^2 - 6(-0.9))/ ((-0.9)^2 -(-0.9) -2).

It's also worth remembering GIGO.

Not too many years ago I'd expect a student to be capable of replacing -0.9 with -9/10 and then calculating an exact value using fractions, without using a calculator. Times change.

Originally Posted by mr fantastic

Here's a thought, old fashioned as it is: Type in ((-0.9)^2 - 6(-0.9))/ ((-0.9)^2 -(-0.9) -2).

It's also worth remembering GIGO.

Not too many years ago I'd expect a student to be capable of replacing -0.9 with -9/10 and then calculating an exact value using fractions, without using a calculator. Times change.

I admit that I rely heavily on my calculator but during testing situations, I need the calculator's accuracy and speed so setting it up to display what I what and when I want is crucial.
Also i understand you can easily plug in the original equation but I was just listing -0.9 as 1 x value out of 14 x value that I had to calculate for. I ended up doing this for the homework anyways because I can not set up my ti83 to show more than 5 decimal places in the graph/table

plugging in every single x value ranging from -0.5 through -1.001 into the original equation might not be the most effective way to work during a 1-2 hour testing period.

Originally Posted by NeoSonata

I admit that I rely heavily on my calculator but during testing situations, I need the calculator's accuracy and speed so setting it up to display what I what and when I want is crucial.
Also i understand you can easily plug in the original equation but I was just listing -0.9 as 1 x value out of 14 x value that I had to calculate for. I ended up doing this for the homework anyways because I can not set up my ti83 to show more than 5 decimal places in the graph/table

plugging in every single x value ranging from -0.5 through -1.001 into the original equation might not be the most effective way to work during a 1-2 hour testing period.

The problem is not only the way you are using your calculator but what you think the answer should look like. Six decimal places means that you should have six figures after the decimal point even if the trailing digits are zeros.

CB

Originally Posted by NeoSonata

My ti83 will not show the 6 decimal place when I input it into the table or trace it on the graph. any thoughts?

Learn how to use your calculator! This calculation can be entered at the keyboard in normal calculator mode and should display something between 12 and 14 significant digits in the output which is more than enough to deal with this question.

CB

Originally Posted by mr fantastic

Here's a thought, old fashioned as it is: Type in ((-0.9)^2 - 6(-0.9))/ ((-0.9)^2 -(-0.9) -2).

It's also worth remembering GIGO.

Not too many years ago I'd expect a student to be capable of replacing -0.9 with -9/10 and then calculating an exact value using fractions, without using a calculator. Times change.

Then finish their homework and take the dinosaur for a walk

CB

Originally Posted by NeoSonata

Evaluate the function at the given numbers (correct to six decimal places).

f(x) = (x^2 - 6x)/ (x^2 -x -2)


x= -0.9 y: -21.41

Webassign marked this as incorrect. How is this incorrect? I plug that exact value into the the equation on my ti 83 and got that answer. Webassign also marked me wrong for other decimal values except for
0, -.5, and 2

by long division equals to:

Hence:



So:



By the way what is the latex code for the long division?

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1200000 | 29



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Originally Posted by NeoSonata

Evaluate the function at the given numbers (correct to six decimal places).

f(x) = (x^2 - 6x)/ (x^2 -x -2)

[snip]

Webassign also marked me wrong for other decimal values except for
0, -.5, and 2

I can find the exact value of y for these values of x (and x = 0 is particularly simple, by the way) faster using mental arithmetic than finding, turning on and using a calculator. And it is very simple to work using pen and paper .....

I'd argue that a student with a competent grasp of basic arithmetic should be able to as well, at least for x = 2 .....

It's well and good to use a calculator, but like any tool, the calculator is only as effective as the mind that is directing it.