To determine $f(1 - a)$, we substitute $1 - a$ into the function $f(x) = x^2 + 2x - 3$.

So, we replace $x$ with $1 - a$:

$f(1 - a) = (1 - a)^2 + 2(1 - a) - 3$

Now, let's simplify:

1. Expand $(1 - a)^2$: $(1 - a)^2 = 1^2 - 2(1)(a) + a^2 = 1 - 2a + a^2$

2. Expand $2(1 - a)$: $2(1 - a) = 2 - 2a$

Now, putting everything together:

$f(1 - a) = (1 - 2a + a^2) + (2 - 2a) - 3$

Simplify the expression:

$f(1 - a) = 1 - 2a + a^2 + 2 - 2a - 3$

Combine like terms:

$f(1 - a) = a^2 - 4a + 0$

So, the simplified expression for $f(1 - a)$ is:

$f(1 - a) = a^2 - 4a$

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Would you like further clarification on this process or have any additional questions?

Related Questions:

1. How would we determine the value of $f(x)$ for $x = 2$?
2. What is the vertex of the quadratic function $f(x) = x^2 + 2x - 3$?
3. Can we find the inverse of the function $f(x) = x^2 + 2x - 3$?
4. What happens to the graph of $f(x) = x^2 + 2x - 3$ if we add a constant to $x$?
5. How do we find the roots of the quadratic function $f(x) = x^2 + 2x - 3$?

Tip:

When dealing with quadratic functions, remember to use the distributive property and combine like terms carefully to simplify expressions!