Let's work through this step-by-step:
First, we need to solve the quadratic equation [tex]\( f(x) = x^2 - 2x - 3 \)[/tex] for [tex]\( x \)[/tex] to find its roots. These roots are the values of [tex]\( x \)[/tex] where [tex]\( f(x) = 0 \)[/tex].
To solve [tex]\( x^2 - 2x - 3 = 0 \)[/tex], we can factor the quadratic equation:
[tex]\[ x^2 - 2x - 3 = (x - 3)(x + 1) = 0 \][/tex]
Setting each factor to zero gives us:
[tex]\[ x - 3 = 0 \][/tex]
[tex]\[ x = 3 \][/tex]
and
[tex]\[ x + 1 = 0 \][/tex]
[tex]\[ x = -1 \][/tex]
So, the solutions to the quadratic equation [tex]\( x^2 - 2x - 3 = 0 \)[/tex] are [tex]\( x = -1 \)[/tex] and [tex]\( x = 3 \)[/tex].
Next, let's evaluate the function [tex]\( f(x) \)[/tex] at the given [tex]\( x \)[/tex] values to fill in the table. Specifically, we want to know [tex]\( f(x) \)[/tex] at [tex]\( x = -1, 0, 1, 2, \)[/tex] and [tex]\( 3 \)[/tex].
1. For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = (-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0 \][/tex]
2. For [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = (0)^2 - 2(0) - 3 = -3 \][/tex]
3. For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = (1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4 \][/tex]
4. For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = (2)^2 - 2(2) - 3 = 4 - 4 - 3 = -3 \][/tex]
5. For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = (3)^2 - 2(3) - 3 = 9 - 6 - 3 = 0 \][/tex]
Now, we can fill in the table:
[tex]\[
\begin{array}{cr|r}
f(x)=x^2-2x-3 & x & y \\
(-1, [0]) ; ([3], [0]) & 0 & -3 \\
& 1 & -4 \\
2 & -3 \\
& 3 & 0
\end{array}
\][/tex]
- When [tex]\( x = -1 \)[/tex], [tex]\( f(x) = 0 \)[/tex].
- When [tex]\( x = 0 \)[/tex], [tex]\( f(x) = -3 \)[/tex].
- When [tex]\( x = 1 \)[/tex], [tex]\( f(x) = -4 \)[/tex].
- When [tex]\( x = 2 \)[/tex], [tex]\( f(x) = -3 \)[/tex].
- When [tex]\( x = 3 \)[/tex], [tex]\( f(x) = 0 \)[/tex].
Thus, the complete solution to the given question has been detailed through step-by-step evaluations and factorization of the quadratic equation.
