$$kx+y=3$$
$$y=-x^2+k$$
In the system of equations above, \(k\) is a constant. When the equations are graphed in the \(xy\)-plane, the graphs intersect at exactly one point. What is a possible value of \(k\)?
See System of Equations for a review of systems.
We first substitute the value of \(y\) to combine the equations.
$$
kx+\colorbox{aqua}{$y$}=3$$
$$
y=-x^2+k
$$
$$ kx\colorbox{aqua}{$-x^2+k$}=3 $$
Written in standard form:
$$ -x^2+kx+k-3=0 $$
We can inspect the discriminant of the quadratic (the portion under the square root of the quadratic formula):
$$ \text{Quadratic Formula} = \frac{-b\pm\sqrt{\colorbox{aqua}{$b^2-4ac$}}}{2a} $$
$$ \text{Discriminant} = b^2-4ac$$
The cases for the discriminant are as follows:
- \(D\gt0\): 2 real solutions
- \(D=0\): 1 real solutions
- \(D\lt0\): 0 real solutions
$$ \colorbox{aqua}{$-$}x^2+\colorbox{yellow}{$k$}x+\colorbox{chartreuse}{$k-3$}=0 $$
$$ \text{Discriminant}= k^2-4(-1)(k-3) $$
Since we want the discriminat to be zero for 1 solution:
$$ 0 = k^2+4k-12=0 $$
$$ 0= (k+6)(k-3) $$
$$ k=-6 \text{ and } 3 $$