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

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

fskilnik · Answer

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

? = k + p

y = {x^2} + mx + \left( {8 - m} ight)\,\,\,\,\,\,,\,\,\,m\,\,{	ext{cte}}\,\,\,\,\,\left( * ight)

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

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

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

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

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

Reminder:
 \left( \odot ight)\,\,y = a{x^2} + bx + c\,\,\,\,\left( {a 
e 0} ight)\,\,\,\, \Rightarrow \,\,\,\,\,\,{x_{{	ext{vertex}}}} = - \frac{b}{{2a}}

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

Regards,
fskilnik.

amascarenhas · Answer

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

amascarenhas · Answer

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

fskilnik · Answer

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.)

Thank you for your contribution!

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.

fskilnik · Answer

Let me present a second possible solution for the interested readers:

? = k + p

y = {x^2} + mx + \left( {8 - m} ight)\,\,\,\,\,\,,\,\,\,m\,\,{	ext{cte}}\,\,\,\,\,\left( * ight)

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

0\mathop > \limits^{{	ext{figure}}} k = {x_{{	ext{vertex}}}}\mathop = \limits^{\left( \odot ight)} \,\, - \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} ight)\,\,\, \in \,\,\,{	ext{graph}} \hfill \
 m = 4 \hfill \ 
\end{gathered} ight\}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * ight)} \,\,\,\,p = \,\,{0^2} + 4 \cdot 0 + \left( {8 - 4} ight)\,\,\,\,\, \Rightarrow \,\,\,\,\,\,p = 4

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

Reminder:
 \left( \odot ight)\,\,y = a{x^2} + bx + c\,\,\,\,\left( {a 
e 0} ight)\,\,\,\, \Rightarrow \,\,\,\,\,\,{x_{{	ext{vertex}}}} = - \frac{b}{{2a}}

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

Regards,
fskilnik.

GMATGuruNY · Answer

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.

B

anse · Answer

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 
Hence 4+(-2) = 2 (Answer)

Kezia9 · Answer

Hi,

Thank you for the explanation. How can we assume a value for m?????

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

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

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

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

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards