The problem involves finding the value of $m$ such that the vertex (top) of the parabola $y = f(x)$ lies on the $x$-axis. The given function is:

$f(x) = x^2 - (m+2)x + 2m$

Step-by-step solution:

1. Vertex formula:

The vertex of a parabola $ax^2 + bx + c$ has an $x$-coordinate given by: $x_v = -\frac{b}{2a}$

Here, $a = 1$, $b = -(m+2)$, and $c = 2m$. Substituting $b = -(m+2)$: $x_v = -\frac{-(m+2)}{2 \cdot 1} = \frac{m+2}{2}$

2. Substitute $x_v$ into $f(x)$ to find $y_v$:

The $y$-coordinate of the vertex is obtained by substituting $x_v$ back into the equation for $f(x)$. We want $y_v = 0$ (the vertex lies on the $x$-axis). Substituting $x = \frac{m+2}{2}$ into $f(x)$: $f\left(\frac{m+2}{2}\right) = \left(\frac{m+2}{2}\right)^2 - (m+2)\left(\frac{m+2}{2}\right) + 2m$

3. Simplify $f(x_v) = 0$:

Expanding $f\left(\frac{m+2}{2}\right)$: $f\left(\frac{m+2}{2}\right) = \frac{(m+2)^2}{4} - \frac{(m+2)^2}{2} + 2m$

Combine terms with a common denominator: $f\left(\frac{m+2}{2}\right) = \frac{(m+2)^2}{4} - \frac{2(m+2)^2}{4} + 2m$ $f\left(\frac{m+2}{2}\right) = \frac{(m+2)^2 - 2(m+2)^2}{4} + 2m$ $f\left(\frac{m+2}{2}\right) = \frac{- (m+2)^2}{4} + 2m$

Set $f(x_v) = 0$: $\frac{-(m+2)^2}{4} + 2m = 0$

4. Solve for $m$:

Multiply through by 4 to eliminate the fraction: $-(m+2)^2 + 8m = 0$ $-(m^2 + 4m + 4) + 8m = 0$ $-m^2 - 4m - 4 + 8m = 0$ $-m^2 + 4m - 4 = 0$

Multiply through by $-1$: $m^2 - 4m + 4 = 0$

Factorize the quadratic: $(m-2)^2 = 0$

So: $m = 2$

Final Answer:

The value of $m$ is: $\boxed{m = 2}$

Let me know if you would like further clarification or a breakdown of specific steps.

Related Questions:

1. What is the general condition for a parabola's vertex to lie on the $x$-axis?
2. How does the coefficient of $x^2$ affect the parabola's shape?
3. Can $m$ take any other values to satisfy this problem?
4. What happens if $m = -2$ in this function?
5. How can we confirm the vertex coordinate algebraically?

Tip:

When solving quadratic equations, always check for opportunities to factorize before resorting to the quadratic formula for efficiency.