Algebra I Exam Notes

Function Transformations:
- $g(x) = f(x-3)$ represents a translation to the right.
- $h(x) = f(x)$ implies no transformation, so it remains unchanged.
- $j(x) = f(x) + 9$ represents a translation upward.

The expression $2x + (-7)^2$ is equivalent to $2x + bx + 49$ .
To find the value of $b$ , we need to compare the given expressions.
Since $2x + (-7)^2 = 2x + 49$ and it's equivalent to $2x + bx + 49$ , then $b = 0$ .

Given the system of equations:
- $2x - 3y = 7$
- $6x - 9y = 21$
Notice that the second equation is just 3 times the first equation. This means the two equations are linearly dependent and represent the same line.
Therefore, the system has infinitely many solutions.

The equation $x^2 - 8x - 5 = 0$ can be transformed into the form $(x - p)^2 = q$ .
To complete the square:
1. Take half of the coefficient of the $x$ term: $\frac{-8}{2} = -4$ .
2. Square the result: $(-4)^2 = 16$ .
3. Add and subtract this value in the equation: $x^2 - 8x + 16 - 16 - 5 = 0$ .
4. Rewrite as a squared term: $(x - 4)^2 - 21 = 0$ .
5. So, $(x - 4)^2 = 21$ .
Thus, $p = 4$ and $q = 21$ .

The expression $(ax + b)(ax - b)$ represents a difference of squares, which expands to $a^2x^2 - b^2$ .
We need to identify which of the given options fits this form:
- A. $x^2 - 9 = (x + 3)(x - 3)$ , where $a = 1$ and $b = 3$ .
- B. $x^2 - 11$ cannot be written in the form $(ax + b)(ax - b)$ where a and b are integers, since 11 is not a perfect square.
- C. $4x^2 - 1 = (2x + 1)(2x - 1)$ , where $a = 2$ and $b = 1$ .
- D. $4x^2 - 2$ cannot be written in the form $(ax + b)(ax - b)$ where a and b are integers, since 2 is not a perfect square.
- E. $9x^2 - 4 = (3x + 2)(3x - 2)$ , where $a = 3$ and $b = 2$ .
Therefore, the expressions that can be written in the given form are A, C, and E.