Solve y=x^2-|0x+1|

$$y = x ^ { 2 } - | 0 x + 1 |$$

Solve for y

$$y=x^{2}-1$$

Solution Steps

Anything times zero gives zero.

$$y=x^{2}-|0+1|$$

Add 0 and 1 to get 1.

$$y=x^{2}-|1|$$

The absolute value of a real number a is a when a\geq 0, or -a when a<0. The absolute value of 1 is 1.

$$y=x^{2}-1$$

Solve for x (complex solution)

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

$$x=\sqrt{y+1}$$

Solve for x

$$x=\sqrt{y+1}$$

$$x=-\sqrt{y+1}\text{, }y\geq -1$$