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# In the xy-coordinate plane, the graph of y=x^2-kx-6, where

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In the xy-coordinate plane, the graph of y=x^2-kx-6, where [#permalink]
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Edit: useful vocabulary to know ... "crosses the x-axis at..." = "x-intercepts" = "roots" = "solutions" = "zeroes"
https://www.purplemath.com/modules/intrcept.htm

Originally posted by energetics on 01 Jun 2019, 09:19.
Last edited by energetics on 01 Jun 2019, 10:53, edited 1 time in total.
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Re: In the xy-coordinate plane, the graph of y=x^2-kx-6, where [#permalink]
I am not sure what you meant.

My two cents ,

But , if a graph is passing through (a,b), then this point will satisfy the graph equation

means if we out x = a and y = b in the original graph equation , then

LHS = RHS

using same logic, here we found the value of K by putting y =0 and x =1 in the original graph equation .

Hope this helps.

Posted from my mobile device
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In the xy-coordinate plane, the graph of y=x^2-kx-6, where [#permalink]
Hi Bunuel

why do you removed the K in that calculation in bold, how can I recognize that I have to do so?

Quote:
If y=(x+2)(x−3)=x2−x−6y=(x+2)(x−3)=x2−x−6, so if the x-intercepts, are -2 and 3, then k=1;
If y=(x−2)(x+3)=x2+x−6y=(x−2)(x+3)=x2+x−6, so if the x-intercepts are 2 and -3, then k=-1.

Many thanks
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Re: In the xy-coordinate plane, the graph of y=x^2-kx-6, where [#permalink]
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gastoneMIT wrote:
Hi Bunuel

why do you removed the K in that calculation in bold, how can I recognize that I have to do so?

Quote:
If y=(x+2)(x−3)=x2−x−6y=(x+2)(x−3)=x2−x−6, so if the x-intercepts, are -2 and 3, then k=1;
If y=(x−2)(x+3)=x2+x−6y=(x−2)(x+3)=x2+x−6, so if the x-intercepts are 2 and -3, then k=-1.

Many thanks

We need to find the value of k. For (1) the solution above demonstrates that k could take more than one value, which indicates that (1) is not sufficient to find the single numerical value of k. For example, if $$y=(x+2)(x-3)=x^2-x-6$$, (the x-intercepts, are integers -2 and 3), then k=1 but if $$y=(x-2)(x+3)=x^2+x-6$$, (the x-intercepts are integers 2 and -3), then k=-1.
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Re: In the xy-coordinate plane, the graph of y=x^2-kx-6, where [#permalink]
Is this question really sub-600 level? It takes a bit of brian to split and find k's possible values for roots to be integers.
Bunuel any thoughts?
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Re: In the xy-coordinate plane, the graph of y=x^2-kx-6, where [#permalink]
kittle wrote:
Is this question really sub-600 level? It takes a bit of brian to split and find k's possible values for roots to be integers.
Bunuel any thoughts?

The difficulty level of a question is calculated automatically based on the timer stats from the users who attempted the question.
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Re: In the xy-coordinate plane, the graph of y=x^2-kx-6, where [#permalink]
In the xy-coordinate plane, the graph of $$y=x^2-kx-6$$, where k is a constant, crosses the x-axis at two points. What is the value of k?

(1) For each of the two points where the graph crosses the x-axis, the x-coordinate is an integer.
(2) The graph crosses the x-axis at (1,0).
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Re: In the xy-coordinate plane, the graph of y=x^2-kx-6, where [#permalink]
Bunuel wrote:
tr1 wrote:
Hey everybody,

this forum has helped me immensely in analyzing my wrong answers on the GMATprep tests. I have not found an explained answer to this question yet, so here we go:

In the xy-coordinate plane, the graph of $$y=x^2-kx-6$$, where k is a constant, crosses the x-axis at two points. What is the value of k?

(1) For each of the two points where the graph crosses the x-axis, the x-coordinate is an integer.
(2) The graph crosses the x-axis at (1,0).

Thanks,
Tim

In the xy-coordinate plane, the graph of $$y=x^2-kx-6$$, where k is a constant, crosses the x-axis at two points. What is the value of k?

(1) For each of the two points where the graph crosses the x-axis, the x-coordinate is an integer.

If $$y=(x+2)(x-3)=x^2-x-6$$, so if the x-intercepts, are -2 and 3, then k=1;
If $$y=(x-2)(x+3)=x^2+x-6$$, so if the x-intercepts are 2 and -3, then k=-1.

Not sufficient.

You could notice that $$y=x^2-kx-6$$ can be factored in many ways so that x-intercepts are integers:
$$y=(x+2)(x-3)$$,
$$y=(x-2)(x+3)$$,
$$y=(x+6)(x-1)$$,
$$y=(x-6)(x+1)$$,
...

(2) The graph crosses the x-axis at (1,0) --> substitute x=1 and y=0, into the equation: $$0=1^2-k-6$$ --> k=-5. Sufficient.

Hi Bunuel, given equation is $$y=x^2-kx-6$$, so not quite sure that how you can use different equation in St 1 as $$y =x^2+x-6$$ is not comply with $$y=x^2-kx-6$$?