To solve this problem, we need to find the value of $m$ such that the curve of the function $f(x) = x^2 + 10x + (m - 1)$ is tangent to the x-axis.

For a quadratic function $ax^2 + bx + c$, the condition for the curve to be tangent to the x-axis is that the discriminant $\Delta$ equals zero. The discriminant is given by:

$\Delta = b^2 - 4ac$

In this case:

• $a = 1$ (coefficient of $x^2$),
• $b = 10$ (coefficient of $x$),
• $c = m - 1$.

So, we set up the discriminant as follows:

$\Delta = 10^2 - 4 \cdot 1 \cdot (m - 1)$ $0 = 100 - 4(m - 1)$

Solving for $m$:

$0 = 100 - 4m + 4$ $0 = 104 - 4m$ $4m = 104$ $m = 26$

Thus, the possible value of $m$ for the curve to be tangent to the x-axis is $m = 26$.

Would you like further explanation or details about any part of this solution?

Here are 5 related questions for further practice:

1. What happens if the discriminant $\Delta > 0$ in a quadratic function?
2. How many intersection points does a parabola have with the x-axis if $\Delta < 0$?
3. For the function $f(x) = ax^2 + bx + c$, how do you determine the vertex?
4. How does the value of $a$ in $ax^2 + bx + c$ affect the direction of the parabola's opening?
5. If a parabola is tangent to the x-axis at $x = p$, how can you find the value of $p$?

Tip: When solving quadratic equations, always check the discriminant to understand the nature of the roots (real and distinct, real and equal, or complex).