Question
The entire graphs of the equation $$y = {x^2} + kx - x + 9$$ is strictly above the $$x$$-axis if and only if
A.
$$k < 7$$
B.
$$ - 5 < k < 7$$
C.
$$k > - 5$$
D.
None of these.
Answer :
$$ - 5 < k < 7$$
Solution :
$$y = {x^2} + \left( {k - 1} \right)x + 9 = {\left( {x + \frac{{k + 1}}{2}} \right)^2} + 9 - {\left( {\frac{{k - 1}}{2}} \right)^2}$$
For entire graph to be above $$x$$-axis, we should have
$$9 - {\left( {\frac{{k - 1}}{2}} \right)^2} > 0$$
$$\eqalign{ & \Rightarrow {k^2} - 2k - 35 < 0 \Rightarrow \left( {k - 7} \right)\left( {k + 5} \right) < 0 \cr & {\text{i}}{\text{.e,}} - 5 < k < 7 \cr} $$
$$y = {x^2} + \left( {k - 1} \right)x + 9 = {\left( {x + \frac{{k + 1}}{2}} \right)^2} + 9 - {\left( {\frac{{k - 1}}{2}} \right)^2}$$
For entire graph to be above $$x$$-axis, we should have
$$9 - {\left( {\frac{{k - 1}}{2}} \right)^2} > 0$$
$$\eqalign{ & \Rightarrow {k^2} - 2k - 35 < 0 \Rightarrow \left( {k - 7} \right)\left( {k + 5} \right) < 0 \cr & {\text{i}}{\text{.e,}} - 5 < k < 7 \cr} $$