Problem Solving

[GMATH course practice question]

Points (x,y) in the rectangular coordinate plane such that y = x^2 + mx + (8-m) 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

Answer: ___B__

fskilnik wrote:[GMATH course practice question]

Points (x,y) in the rectangular coordinate plane such that y = x^2 + mx + (8-m) 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.

fskilnik wrote: Points (x,y) in the rectangular coordinate plane such that y = x^2 + mx + (8-m) 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

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.

fskilnik wrote:[GMATH course practice question]

Points (x,y) in the rectangular coordinate plane such that y = x^2 + mx + (8-m) 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.

The correct answer is B.