Find the slope of $L$ using the two labeled points, then use one point to write the line in the form $y=mx+n$.

After setting the line equal to the parabola, you get a quadratic in $x$. A quadratic has exactly one real solution when its discriminant equals $0$.

Enter the line: y=0.5x+1.

From the axis of symmetry $x=2$, set $b=-4$ and enter the parabola with a slider for $c$: y=x^2-4x+c.

Adjust the slider for $c$ until the line touches the parabola at exactly one point (you should see a single intersection instead of two).

When there is exactly one intersection point, read the corresponding slider value of $c$ and select the answer choice that matches it.

For $y=x^2+bx+c$, the axis of symmetry is $x=-\frac{b}{2}$.

From the graph, the dashed axis is at $x=2$, so

From the graph, line $L$ passes through $(0,1)$ and $(6,4)$.

Its slope is

so the line is

Substitute $b=-4$ into the parabola: $y=x^2-4x+c$.

At intersection points with the line,

Move everything to one side:

Because the line touches the parabola at exactly one point, the quadratic has exactly one real solution, so its discriminant is $0$:

Compute and solve: