Answer
See explanations.
Work Step by Step
Step 1. List the given equations:
\begin{cases} y=x^2 \\ y=x+k \end{cases}
Step 2. Graph both equations with different $k=0,-0.5,0.25$ values as shown, where $k=0$ gives two intersection points, $k=-0.5$ gives no intersection points, and $k=-0.25$ gives exactly one intersection point.
Step 3. Make a conjecture: the system has two solutions when $k\gt -0.25$, one solution when $k=-0.25$, and no solution when $k\lt -0.25$
Step 4. Refer to the equations in step-1, we have $x^2=x+k$ or $x^2-x-k=0$. Use the quadratic formula,
$x=\frac{1\pm \sqrt {1+4k}}{2}$
Step 5. Clearly, there will be exactly one solution for $x=\frac{1}{2}$ when $k=-\frac{1}{4}=-0.25$, two real solutions when $k\lt -0.25$, and no real solution when $k\gt -0.25$
