Precalc. AB Semester Final Part 1

$$\left(-\infty,\ 1\right]$$

$$\left[-1,\ 3\right]$$

$$\left[3,\ \infty\right)$$

$$\left(-\infty,\ 1\right]\cup\left[3,\ \infty\right]$$

$$\left(-\infty,\ \infty\right)$$

$$\left(\sqrt[]{5},\ \infty\right)$$

$$\left(-\infty,\ 5\right)\cup\left(5,\ \infty\right)$$

$$\left(-\infty,\ 5\right]$$

$$\left(-\infty,\ 5\right)$$

$$\left(-\infty,\ -5\right)\cup\left(-5,\ 5\right)\cup\left(5,\ \infty\right)\$$

$$\left(-\infty,\ -5\right)\cup\left(-5,\ \infty\right)\$$

$$\left(-\infty,\ 5\right)\cup\left(5,\ \infty\right)\$$

$$\left(-\infty,\ 1\right)\cup\left(1,\ \infty\right)\$$

$$\left[7,\ \infty\right)$$

$$\left(-\infty,\ \infty\right)$$

$$\left(-7,\ \infty\right)$$

$$\left(7,\ \infty\right)$$

$$y=\sqrt[]{x+2}-4$$

$$y\ =-\left(x+1\right)^2-4$$

$$y=-\left(x-4\right)^2$$

$$y=-4+\left|x+3\right|$$

$$\left(4,\ \infty\right)$$

$$\left(-\infty,\ 4\right)$$

(-4, 4)

$$y=\sqrt[]{x+2}+4$$

$$y=\sqrt[]{x-2}-4$$

$$y=\sqrt[]{x-2}+4$$

$$y=\sqrt[]{x+2}-4$$

12x² + 9x + 23

4x² + 9x + 7

4x² + 3x - 3

12x² + 9x + 7

$$3x^4-2x^3+4x^2-6x+\frac{11}{x+2}$$

$$3x^4-2x^3+6x^2+\frac{-13}{x+2}$$

$$3x^4-2x^3+4x^2-6x+12+\frac{-25}{x+2}$$

$$3x^4-2x^3+6x^2-12+\frac{23}{x+2}$$

y = 0

y = 8

y = -8

y = 1

y = -1