Solve y=1-(x+1/2)^2

$$y = 1 - ( \frac { x + 1 } { 2 } ) ^ { 2 }$$

Solve for y

$$y=-\frac{\left(x-1\right)\left(x+3\right)}{4}$$

Solution Steps

To raise \frac{x+1}{2} to a power, raise both numerator and denominator to the power and then divide.

$$y=1-\frac{\left(x+1\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$y=1-\frac{x^{2}+2x+1}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$y=1-\frac{x^{2}+2x+1}{4}$$

Divide each term of x^{2}+2x+1 by 4 to get \frac{1}{4}x^{2}+\frac{1}{2}x+\frac{1}{4}.

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

To find the opposite of \frac{1}{4}x^{2}+\frac{1}{2}x+\frac{1}{4}, find the opposite of each term.

$$y=1-\frac{1}{4}x^{2}-\frac{1}{2}x-\frac{1}{4}$$

Subtract \frac{1}{4} from 1 to get \frac{3}{4}.

$$y=\frac{3}{4}-\frac{1}{4}x^{2}-\frac{1}{2}x$$

Solve for x (complex solution)

$$x=-2\sqrt{1-y}-1$$

$$x=2\sqrt{1-y}-1$$

Solve for x

$$x=-2\sqrt{1-y}-1$$

$$x=2\sqrt{1-y}-1\text{, }y\leq 1$$