why is it that when 'using intercepts and symmetry to sketch a graph' does the equation x-y^2=1 get the y replaced w/ a negative [looking like this - x-(-y)^2=1 ] wind up being simplified to x-y^2=1 instead of being simplified to x+y^2=1 ?

Question

why is it that when 'using intercepts and symmetry to sketch a graph' does the equation x-y^2=1 get the y replaced w/ a negative [looking like this ->    x-(-y)^2=1 ] wind up being simplified to x-y^2=1 instead of being simplified to x+y^2=1 ?

dan815 · Answer

what thats messed up forget all about that stuff

anonymous · Answer

aw, well :( unfortunately I can not because it is a fundamental question for me to understand the section but thanks

jhannybean · Answer

$$\large x-(-y)^2 = 1 $$ solve the (-y)^2 first. $$\large (-y^2) = (-y)(-y) = y^2$$now replace (-y)^2 with y^2$$\large x-y^2 = 1$$

jhannybean · Answer

The y doesn't get replaced by a negative, you just have to remember to solve for your (-y)^2 first before simplifying x-(-y)^2 into x+y^2.

anonymous · Answer

thank you, wowww I cant believe I didn't catch that

dan815 · Answer

i said forget all this stuff because i dont like how ur thinking about equations

dan815 · Answer

we dont think about all this stuff, when ur just building a vertex equations, u just wanna kinda build ur equation knowing where the Vertex will be

dan815 · Answer

dont be restrained to these rules, learn to adapt then u will find math easier as u get better at adapting to different methods

anonymous · Answer

okidiki, thank you foor that insight

anonymous · Answer

okidoki, lol