The exercise requires finding the factorized form of second-degree functions based on given conditions.

1. Problem 1:

The curve of the quadratic function $f(x)$ intercepts the x-axis at $x = 4$ and $x = 5$, and passes through the point $(1, 24)$.

• The factorized form of a second-degree function that intercepts the x-axis at $x = 4$ and $x = 5$ is: $f(x) = a(x - 4)(x - 5)$ Now, we need to find the constant $a$ using the fact that $f(1) = 24$. Substitute $x = 1$ into the equation: $24 = a(1 - 4)(1 - 5)$ $24 = a(-3)(-4)$ $24 = 12a$ $a = 2$ Thus, the factorized form is: $f(x) = 2(x - 4)(x - 5)$

2. Problem 2:

The curve of the quadratic function $f(x)$ intercepts the x-axis at $x = -2$, and passes through the point $(-1, -1)$.

• The factorized form of a second-degree function that intercepts the x-axis at $x = -2$ can be written as: $f(x) = a(x + 2)^2$ Since it passes through the point $(-1, -1)$, we substitute $x = -1$ and $f(x) = -1$ into the equation to solve for $a$: $-1 = a(-1 + 2)^2$ $-1 = a(1^2)$ $-1 = a$ Thus, the factorized form is: $f(x) = -(x + 2)^2$

Would you like more details or have any questions?

Here are 5 related questions for further practice:

1. What is the general factorized form of a quadratic function with given x-intercepts $x_1$ and $x_2$?
2. How can you determine the value of $a$ in a quadratic function?
3. How does the sign of $a$ affect the shape of a parabola?
4. What happens if the quadratic function only touches the x-axis at one point?
5. Can you convert the factorized form to standard form and vice versa?

Tip: Always use a known point (other than the x-intercepts) to solve for the coefficient $a$ in a quadratic function.