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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).
(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).
tr1
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.
Answer: B.
Factoring Quadratics: https://www.purplemath.com/modules/factquad.htm
Solving Quadratic Equations: https://www.purplemath.com/modules/solvquad.htm
Hope this helps.
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General Discussion
This clears it up, thanks a lot!
Posts: 292
Updated on: 01 Jun 2019, 09:53
Originally posted by energetics on 01 Jun 2019, 08:19.
Last edited by energetics on 01 Jun 2019, 09:53, edited 1 time in total.
Last edited by energetics on 01 Jun 2019, 09:53, edited 1 time in total.
Edit: useful vocabulary to know ... "crosses the x-axis at..." = "x-intercepts" = "roots" = "solutions" = "zeroes"
https://www.purplemath.com/modules/intrcept.htm
https://www.purplemath.com/modules/intrcept.htm
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
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
Schools: Harvard '24
Posts: 50
Hi Bunuel
why do you removed the K in that calculation in bold, how can I recognize that I have to do so?
Many thanks
why do you removed the K in that calculation in bold, how can I recognize that I have to do so?
Quote:
Many thanks
gastoneMIT
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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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?
Bunuel any thoughts?
kittle
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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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).
(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).
Attachments
159FB13B-3268-43DD-8BAA-AB344F5C7228.jpeg [ 590.97 KiB | Viewed 11643 times ]
Kimberly77
Joined: 16 Nov 2021
Last visit: 07 Sep 2024
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GMAT 1: 450 Q42 V34
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GMAT 1: 450 Q42 V34
Posts: 418
Bunuel
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 \(y=x^2-kx-6\)?
Also not sure how do you get k=-1?
Could you help clarify? Thanks
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