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Do part A - the sketch. I found the it crosses the X axis at X=13/4. Howcome it crosses at the point X=9/4 as well


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Original post by Lilly1234567890
ImageUploadedByStudent Room1454778492.581143.jpg

Do part A - the sketch. I found the it crosses the X axis at X=13/4. Howcome it crosses at the point X=9/4 as well


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It doesn't. ln4(9/4)12=ln30\ln |4(9/4) - 12| = \ln 3 \not{=} 0.
Reply 2
Original post by Lilly1234567890
ImageUploadedByStudent Room1454778492.581143.jpg

Do part A - the sketch. I found the it crosses the X axis at X=13/4. Howcome it crosses at the point X=9/4 as well


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Are you sure you don't mean x = 11/4 ?

And the reason for the other x-intercept is because of the modulus sign.
You can have a negative value for a logarithm because the modulus sign takes the magnitude of the value (i.e. makes it positive)
Reply 3
Original post by Lilly1234567890
ImageUploadedByStudent Room1454778492.581143.jpg

Do part A - the sketch. I found the it crosses the X axis at X=13/4. Howcome it crosses at the point X=9/4 as well


Posted from TSR Mobile


It meets the x-axis when ln4x12=0    4x12=1    4x12=±1\ln |4x-12| = 0 \iff |4x-12| = 1 \iff 4x - 12 = \pm 1.
Reply 4
Original post by Lilly1234567890
ImageUploadedByStudent Room1454778492.581143.jpg

Do part A - the sketch. I found the it crosses the X axis at X=13/4. Howcome it crosses at the point X=9/4 as well


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using transformations is the best approach for this graph, which is truly difficult to sketch
Original post by morgan8002
It doesn't. ln4(9/4)12=ln30\ln |4(9/4) - 12| = \ln 3 \not{=} 0.


It's meant to look something like that


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Original post by Lilly1234567890
It's meant to look something like that


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(edited 8 years ago)
Original post by RMNDK
Are you sure you don't mean x = 11/4 ?

And the reason for the other x-intercept is because of the modulus sign.
You can have a negative value for a logarithm because the modulus sign takes the magnitude of the value (i.e. makes it positive)


your're right! it is 11/4 thanks got it

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(edited 8 years ago)
Original post by Zacken
It meets the x-axis when ln4x12=0    4x12=1    4x12=±1\ln |4x-12| = 0 \iff |4x-12| = 1 \iff 4x - 12 = \pm 1.


thanks got it !

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