2/11 ~ FSQ04 Review Part 2

$$x=\pm\sqrt[]{\frac{y}{5}}-13$$

$$x=\pm\sqrt{\frac{y+10}{7}}$$

$$x=\pm\sqrt[]{y+10}$$

$$x=\pm\sqrt[]{5y-13}$$

$$x=\pm\sqrt[]{\frac{y}{5}}-13$$

$$x=\pm\sqrt{\frac{y+13}{5}}$$

$$x=\pm\sqrt[]{5y-13}$$

$$x=\pm\sqrt[]{\frac{y-13}{5}}$$

$$\left(4,-1\right)$$

$$\left(-4,-1\right)$$

$$\left(-4,1\right)$$

$$\left(4,1\right)$$

$$f\left(x\right)=x^2-4x-12$$

$$f\left(x\right)=x^2+4x-12$$

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

$$f\left(x\right)=x^2-4x-12$$

$$f\left(x\right)=x^2+4x-12$$

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

$$f\left(x\right)=x^2-4x-12$$

$$f\left(x\right)=x^2+4x-12$$

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

$$f(x)=2(x+5)(x+1)$$

$$f(x)=-(x-1)(x-5)$$

$$f(x)=3(x+5)(x+1)$$

$$f(x)=-(x+1)(x+5)$$

$$x^2+8x+240=0$$

$$x^2-8x+240=0$$

$$x^2+8x-240=0$$

$$x^2-8x-240=0$$