IB-why is it that f(x)=2x+1 intercepts y at 0.5,not 1?

Leah5467

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1.So i came to the conclusion that by making the euqation:f(x)=2(x+0.5) can reach the answer. But is my step correct?

2.Also why can't it first multiply by 2-->f(x)=2x,and then the whole f(x)+1? And then it will be intercepting at 1. Or is there something wrong with it?
 

Harry_the_cat

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Are you asking about Q2a(i)?
If f(x) = 2x + 1, then f(-x) = 2(-x) + 1 = -2x +1.

In relation to the title of your post:
f(x) = 2x + 1 does intersect the y-axis at 1.

I'm not exactly sure what your question is
 

Dr.Peterson

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The problem is,

Find f(-x) for f(x) = 2x + 1.​

Please explain why you wrote f(x)=2x and f(x)+1. (I'm guessing that you are not recognizing that f(x) has a specific meaning in this problem, and are just writing "f(x)" to mean "what I've done so far".) It looks like you're trying to solve a different problem than what you showed.

All you have to do for this problem is to replace x in 2x + 1 with (-x).

Now, it is true that 2x + 1 = 2(x + 0.5); but why would you do that in this problem? That is f(x), not f(-x), and it doesn't help toward the answer.Furthermore, the y-intercept of f(x) = 2x + 1 is 1, regardless of which way you write it.
 

HallsofIvy

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The line y= 2x+ 1 intercepts the y-axis at the point (0, 1). It intercepts the x-axis at the point (1/2, 0). You may be confusing the two.
 

Harry_the_cat

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HallsofIvy, typo? x-intercept is (-1/2, 0)
 

Leah5467

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yes,it is x-axis that is 0.5 and -0.5. I typed it wrongly. But i don't really understand what transformation the graph went through to make x-intercept 0.5 and -0.5.
 

Jomo

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Yes, f(x) has an x-int at x=-1/2 and f(-x) has an x-int at x= 1/2
 

Leah5467

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I am learning the transformation of graph and this question appears in the transformation of the graph. But i don't really understand why x-intercept is 0.5 and -0.5 after the transformation of 2x+1. Please help!
 

Harry_the_cat

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OK. Let's step you through this.

f(x) =2x +1 is a straight line with a y-int of 1 and an x-int of -1/2. Correct?
 

Harry_the_cat

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In general, how does the graph of y=f(-x) compare with the graph of y=f(x) for any function? That is, what transformation does the original graph undergo?
 

Leah5467

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I think it reflects on the y-axis.
 

Harry_the_cat

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Yes exactly. So if the original line f(x) = 2x+1 has an x-intercept of -1/2, what is the x-intercept after the line has reflected in the y-axis?
Draw a quick sketch if you need to.
 

Leah5467

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ahh i see. Thank you. Then did f(x)=2x+1 go through any transformation? or the x-intercept is calculated by subbing y as 0?
 

Harry_the_cat

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Yes the line f(x) =2x+1 was reflected in the y-axis.
The x-intercept of f(x)=2x+1 is found by substituting y=0.

The x-intercept of f(-x) can be found in one of 2 ways:
1. f(-x) = 2(-x)+1 = -2x+1 and substitute y=0 to find x-int is 1/2
OR
2. Recognise that f(-x) is a reflection of f(x) in the y-axis, so the original x-int of -1/2 gets reflectd onto the x-int of 1/2.
 

Leah5467

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thank you! now i understand.
 
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