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# Points (x,y) in the rectangular coordinate plane such that

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GMATH Teacher
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Points (x,y) in the rectangular coordinate plane such that  [#permalink]

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03 Sep 2018, 05:09
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Difficulty:

95% (hard)

Question Stats:

35% (03:04) correct 65% (02:48) wrong based on 55 sessions

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[GMATH course practice question]

Points $$(x,y)$$ in the rectangular coordinate plane such that $$\,y = {x^2} + mx + \left( {8 - m} \right)\,$$ are presented in the graph shown, where $$m$$ is constant. What is the value of $$k+p$$ ?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

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03Set18_7m.gif [ 6.22 KiB | Viewed 894 times ]

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Re: Points (x,y) in the rectangular coordinate plane such that  [#permalink]

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03 Sep 2018, 08:19
fskilnik wrote:
[GMATH course practice question]

Points $$(x,y)$$ in the rectangular coordinate plane such that $$\,y = {x^2} + mx + \left( {8 - m} \right)\,$$ are presented in the graph shown, where $$m$$ is constant. What is the value of $$k+p$$ ?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

I am sorry no one contributed (yet), but I hope you all enjoy my solution!

$$? = k + p$$

$$y = {x^2} + mx + \left( {8 - m} \right)\,\,\,\,\,\,,\,\,\,m\,\,{\text{cte}}\,\,\,\,\,\left( * \right)$$

$$k = {x_{{\text{vertex}}}}\mathop = \limits^{\left( \odot \right)} \,\, - \frac{m}{{2 \cdot 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,m = - 2k\,\,\,\left( {**} \right)$$

$$\left( {k,0} \right)\,\,\, \in \,\,\,{\text{graph}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,0 = \,\,{k^2} + mk + \left( {8 - m} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,\,\,0 = - {k^2} + 2k + 8$$

$$0 = - {k^2} + 2k + 8\,\,\,\,\mathop \Rightarrow \limits^{{\text{Sum}}\, = \,2\,\,,\,\,\,{\text{Product}}\, = \,\, - 8} \,\,\,\,k = - 2\,\,\,{\text{or}}\,\,\,k = 4\,\,\,\mathop \Rightarrow \limits^{k\, < \,\,0\,\,\,\left( {f{\text{igure}}} \right)} \,\,\,\,k = - 2$$

$$\left( {0,p} \right)\,\,\, \in \,\,\,{\text{graph}}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,p = \,\,{0^2} + m \cdot 0 + \left( {8 - m} \right)\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,\,\,\,\,p = 8 + 2k = 8 + 2\left( { - 2} \right) = 4$$

$$? = k + p = - 2 + 4 = \boxed2$$

Reminder:
$$\left( \odot \right)\,\,y = a{x^2} + bx + c\,\,\,\,\left( {a \ne 0} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\,{x_{{\text{vertex}}}} = - \frac{b}{{2a}}$$

The above follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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03Set18_7m.gif [ 6.22 KiB | Viewed 842 times ]

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Re: Points (x,y) in the rectangular coordinate plane such that  [#permalink]

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03 Sep 2018, 10:14
3
let's say that m = 4

y = sqr(x)+4x+2 =sqr (x+2)

when y = 0 x=-2 (k)

when x=0 y =4 (p)

k+p =4-2 = 2 Anwser B
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Re: Points (x,y) in the rectangular coordinate plane such that  [#permalink]

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03 Sep 2018, 10:15
let's say that m = 4

y = sqr(x)+4x+2 =sqr (x+2)

when y = 0 x=-2 (k)

when x=0 y =4 (p)

k+p =4-2 = 2 Anwser B
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Points (x,y) in the rectangular coordinate plane such that  [#permalink]

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03 Sep 2018, 10:57
amascarenhas wrote:
let's say that m = 4

y = sqr(x)+4x+2 =sqr (x+2)

when y = 0 x=-2 (k)

when x=0 y =4 (p)

k+p =4-2 = 2 Anwser B

Hi, amascarenhas!

I am sure you have meant x^2 in the place of sqr(x) and (x+2)^2 in the place of sqr(x+2) ...

The fact that m = 4 makes the expression y = (x+2)^2 DOES guarantee that the x_vertex of the parabola is -2 (=k), exactly the opposite of half the value of this value of m, what we know (a posteriori) is the correct relationship between these two.

More than that, y_vertex will be zero (for x=-2), exactly as the figure suggests. (The vertex of the parabola must be on the x-axis.)

Regards,
fskilnik.

P.S.: let me be more clear. If you had chosen (say) m=2, you would find the x_vertex = -1 (negative, fine) but the y_vertex would be 5, therefore the vertex would be point (-1,5) , that violates the question stem (figure). In other words, m=4 is not "free" as we wished, therefore exploring "any" particular case would not be possible here.
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Re: Points (x,y) in the rectangular coordinate plane such that  [#permalink]

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03 Sep 2018, 11:29
Let me present a second possible solution for the interested readers:

$$? = k + p$$

$$y = {x^2} + mx + \left( {8 - m} \right)\,\,\,\,\,\,,\,\,\,m\,\,{\text{cte}}\,\,\,\,\,\left( * \right)$$

$$0\mathop = \limits^{{\text{single}}\,\,{\text{root}}} \Delta = {m^2} - 4\left( {8 - m} \right)\,\,\,\,\, \Rightarrow \,\,\,\,{m^2} + 4m - 32 = 0\,\,\,\mathop \Rightarrow \limits^{{\text{Sum}}\, = \, - 4\,\,,\,\,\,{\text{Product}}\, = \,\, - 32} \,\,\,\,m = - 8\,\,\,{\text{or}}\,\,\,m = 4$$

$$0\mathop > \limits^{{\text{figure}}} k = {x_{{\text{vertex}}}}\mathop = \limits^{\left( \odot \right)} \,\, - \frac{m}{{2 \cdot 1}}\,\,\,\,\, \Rightarrow \,\,\,\,\,m > 0\,\,\,\,\, \Rightarrow \,\,\,\,\,m = 4\,\,\,\,\,\mathop \Rightarrow \limits^{k = - m/2} \,\,\,\,\,k = - 2$$

$$\left. \begin{gathered} \left( {0,p} \right)\,\,\, \in \,\,\,{\text{graph}} \hfill \\ m = 4 \hfill \\ \end{gathered} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,p = \,\,{0^2} + 4 \cdot 0 + \left( {8 - 4} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,p = 4$$

$$? = k + p = - 2 + 4 = \boxed2$$

Reminder:
$$\left( \odot \right)\,\,y = a{x^2} + bx + c\,\,\,\,\left( {a \ne 0} \right)\,\,\,\, \Rightarrow \,\,\,\,\,\,{x_{{\text{vertex}}}} = - \frac{b}{{2a}}$$

The above follows the notations and rationale taught in the GMATH method.

Regards,
fskilnik.
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Re: Points (x,y) in the rectangular coordinate plane such that  [#permalink]

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03 Sep 2018, 17:18
1
fskilnik wrote:
[GMATH course practice question]

Points $$(x,y)$$ in the rectangular coordinate plane such that $$\,y = {x^2} + mx + \left( {8 - m} \right)\,$$ are presented in the graph shown, where $$m$$ is constant. What is the value of $$k+p$$ ?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Since the graph intercepts the x-axis to the left of the y-axis and has only one solution, its graph is as follows:
y = (x-k)², where k<0.
The equation above has an x-intercept at (k, 0), as shown in the figure above.

The answer choices represent the sum of k (the x-intercept) and p (the y-intercept).
Since the answer choices are very small, test negative values for k that are close to 0.

Case 1: k=-1
Plugging k=-1 into y = (x-k)², we get:
y = (x-(-1))² = (x+1)² = x² + 2x + 1.
Since the prompt indicates that y = x² + mx + (8-m), Case 1 implies that m=2 and that 8-m = 1.
The equations in red contradict each other, implying that Case 1 is not viable.

Case 2: k=-2
Plugging k=-2 into y = (x-k)², we get:
y = (x-(-2))² = (x+2)² = x² + 4x + 4.
Since the prompt indicates that y = x² + mx + (8-m), Case 2 implies that m=4 and that 8-m = 4.
The equations in green are both viable for m=4, implying that Case 2 is correct and that the equation of the graph is y = (x+2)².

Since y=4 when x=0, the value of p = 4.
Thus, k+p = -2 + 4 = 2.

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Points (x,y) in the rectangular coordinate plane such that  [#permalink]

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Updated on: 05 Sep 2018, 19:10
Hi,

When we look at the equation y=x^2+mx+(8−m) and at the graph we see two things:

1) The equation has only one root.
2) At x=0 , it has a positive intercept i.e. (8-m) > 0

Now When we have only one root ( which are also equal and real ) then according to quadratic equations the determinant shall be 0 (B^2-4AC=0)
so solve for m^2 - 4(8-m) = 0

We shall get two values -8, 4 but these should support the 2nd point above. Hence we conclude that m=4.
On placing the value of m in the equation we get x^2 + 4x + 4 which is (x +2 )^2 and hence x=-2

k is the value of x when y=0 so in other terms k is the root of the equation hence k=-2
p is the y intercept when x=0 i.e. p = ( 8- m ) = 4

Originally posted by anse on 05 Sep 2018, 18:32.
Last edited by anse on 05 Sep 2018, 19:10, edited 1 time in total.
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Re: Points (x,y) in the rectangular coordinate plane such that  [#permalink]

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05 Sep 2018, 18:59
amascarenhas wrote:
let's say that m = 4

y = sqr(x)+4x+2 =sqr (x+2)

when y = 0 x=-2 (k)

when x=0 y =4 (p)

k+p =4-2 = 2 Anwser B

Hi,

Thank you for the explanation. How can we assume a value for m?????
Re: Points (x,y) in the rectangular coordinate plane such that   [#permalink] 05 Sep 2018, 18:59
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