The problem asks to find the value of $f(1)$, given the following conditions:

1. $f(x) = ax^2 + bx - 5$, where $a$ and $b$ are constants.
2. $f(x-1) - f(x) = 4x - 3$ for all real numbers $x$.

Step 1: Express $f(x-1)$

From $f(x) = ax^2 + bx - 5$, let's find $f(x-1)$:

$f(x-1) = a(x-1)^2 + b(x-1) - 5$ Expand the terms:

$f(x-1) = a(x^2 - 2x + 1) + b(x-1) - 5$ $f(x-1) = ax^2 - 2ax + a + bx - b - 5$

Step 2: Compute $f(x-1) - f(x)$

Now, let's subtract $f(x)$ from $f(x-1)$:

$f(x-1) - f(x) = (ax^2 - 2ax + a + bx - b - 5) - (ax^2 + bx - 5)$ Simplify the expression:

$f(x-1) - f(x) = ax^2 - 2ax + a + bx - b - 5 - ax^2 - bx + 5$ $f(x-1) - f(x) = -2ax + a - b$

Step 3: Set the expression equal to $4x - 3$

We know that $f(x-1) - f(x) = 4x - 3$. So,

$-2ax + a - b = 4x - 3$

Step 4: Solve for $a$ and $b$

To find $a$ and $b$, compare the coefficients of $x$ and the constant terms on both sides.

• For the $x$-terms: $-2a = 4 \quad \Rightarrow \quad a = -2$

• For the constant terms: $a - b = -3 \quad \Rightarrow \quad -2 - b = -3 \quad \Rightarrow \quad b = 1$

Step 5: Find $f(1)$

Now that we know $a = -2$ and $b = 1$, we can substitute these values into the expression for $f(x)$:

$f(x) = -2x^2 + x - 5$

To find $f(1)$:

$f(1) = -2(1)^2 + 1 - 5 = -2 + 1 - 5 = -6$

Thus, the value of $f(1)$ is $\boxed{-6}$.

Would you like a detailed explanation of any step or have further questions?

Here are five related questions for further exploration:

1. How would the result change if the equation for $f(x-1) - f(x)$ were different?
2. Can you generalize this method for any quadratic function?
3. How would the solution change if the right-hand side of the equation were $4x - c$, where $c$ is a constant?
4. What is the significance of the relationship $f(x-1) - f(x)$?
5. How would you approach this if the equation for $f(x)$ was cubic instead of quadratic?

Tip: Always expand and simplify both sides of an equation when working with functional relationships.