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bonez
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14 Jan 2009, 8:42 pm

suppose that the x-intercepts of the graph of y=f(x) are -8 and 1. What are the x-intercepts of the graph of y=f(x+4)? and what about the graph of y=2f(x) ?



pakled
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14 Jan 2009, 9:02 pm

I haven't seen equations like this since 1975...;)

There should be any number of free graphing calculator programs out there that can do this. Try Sourceforge, Nonags, or something like that.



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14 Jan 2009, 9:22 pm

bonez wrote:
suppose that the x-intercepts of the graph of y=f(x) are -8 and 1. What are the x-intercepts of the graph of y=f(x+4)? and what about the graph of y=2f(x) ?


Don't hold me to this, but for y=f(x+4), the intercepts will be 4 less than the original x values, so -12 and -3. For the second one, I don't think multiplying it by two will change the x-intercepts, so it's still -8 and 1.



bonez
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14 Jan 2009, 9:28 pm

DNForrest wrote:
bonez wrote:
suppose that the x-intercepts of the graph of y=f(x) are -8 and 1. What are the x-intercepts of the graph of y=f(x+4)? and what about the graph of y=2f(x) ?


Don't hold me to this, but for y=f(x+4), the intercepts will be 4 less than the original x values, so -12 and -3. For the second one, I don't think multiplying it by two will change the x-intercepts, so it's still -8 and 1.
so for y=f(x-3) it would be -5 and 4 ?



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14 Jan 2009, 9:33 pm

bonez wrote:
DNForrest wrote:
bonez wrote:
suppose that the x-intercepts of the graph of y=f(x) are -8 and 1. What are the x-intercepts of the graph of y=f(x+4)? and what about the graph of y=2f(x) ?


Don't hold me to this, but for y=f(x+4), the intercepts will be 4 less than the original x values, so -12 and -3. For the second one, I don't think multiplying it by two will change the x-intercepts, so it's still -8 and 1.
so for y=f(x-3) it would be -5 and 4 ?


Indeed.



Shadow50
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14 Jan 2009, 9:55 pm

DNForrest wrote:
bonez wrote:
suppose that the x-intercepts of the graph of y=f(x) are -8 and 1. What are the x-intercepts of the graph of y=f(x+4)? and what about the graph of y=2f(x) ?


Don't hold me to this, but for y=f(x+4), the intercepts will be 4 less than the original x values, so -12 and -3. For the second one, I don't think multiplying it by two will change the x-intercepts, so it's still -8 and 1.


I was once a maths teacher. What DNForrest stated is quite correct.

y = f(x+4) moves the graph sideways and therefore changes the intercept points along the x axis.

y = 2f(x) stretches the graph about the x axis but does not move the intercept points.

but note that y = f(2x) does NOT have the same effect as y = 2f(x)


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