After you set the two expressions for $y$ equal, expand $(x - 3)^2$ and collect like terms to form a standard quadratic equation $ax^2 + bx + k = 0$.

For the quadratic you formed, think about what condition on the discriminant $b^2 - 4ac$ guarantees that there is one or zero real solution(s) for $x$.

Once you solve the inequality involving $c$ from the discriminant condition, remember that $c$ must be an integer and choose the smallest integer that satisfies your inequality.

In Desmos, enter y = (x + 10)/2 for the line. Then enter y = (x - 3)^2 + c and let Desmos create a slider for c. This will let you move the parabola up and down.

Move the c slider and watch how many intersection points the line and parabola have. Find the value of c where the line just touches the parabola at exactly one point (tangent); beyond this value there will be no intersections. That boundary value of c is the cutoff where the system changes from two real solutions to at most one.

From the boundary value you saw in Desmos, identify the smallest integer value of c that is at or above this cutoff. That integer is the least possible value of c for which the system has at most one real solution.

Rewrite the linear equation in slope-intercept form:

Now both equations are written as $y =$ something:

Set these equal to each other to find the $x$-values where the graphs intersect:

Expand $(x - 3)^2$ and move everything to one side:

Substitute this into the equation:

Bring all terms to the left side:

Combine like terms:

To avoid decimals, multiply the entire equation by $2$:

For a quadratic equation of the form $ax^2 + bx + k = 0$, the discriminant is

There are:

• Two real solutions if $D > 0$,
• One real solution (a repeated root) if $D = 0$,
• No real solutions if $D < 0$.

"At most one real solution" means $D \le 0$.

In our equation $2x^2 - 13x + (8 + 2c) = 0$, we have:

• $a = 2$,
• $b = -13$,
• $k = 8 + 2c$.

So the discriminant is

Simplify this:

Now impose the condition $D \le 0$:

Solve the inequality

Add $16c$ to both sides and subtract $0$:

Divide both sides by $16$:

Compute the fraction:

So $c$ must be at least $6.5625$. Because $c$ is an integer, the smallest integer $c$ that satisfies $c \ge 6.5625$ is $\boxed{7}$. This is the least possible value of $c$ for which the system has at most one real solution.