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is this a calculus question?

how do you figure

so the slope of the normal line is -1/(2x)??

yep

i thought the slope had to be a number

like so that's equal to a slope of -1/2?

ohh for real ok

did you get to this point of the problem: y = c- 1/2?

i don't know i mean y-c = -(1/2x)(x-0)? or -(1/2)(x-0) ???

...the second right

yes... you got it now just simplify the right side to -1/2*x

right obviously so ok then yeah y = c-1/2*x

sorry, i mean -1/2

no x's

wait what

simplifying the right side, the x's cancell out right? you're left with only -1/2

where'd the x go

oh ok so it IS -1/2*x(x-0) right to begin with

just to be more clear, the right side is (-1/(2x))*(x-0)

right right that's what i meant ok

thanks

np

so if c = 1/2, y = 0, and... how does that mean that there is one line and that it's the y-axis?

from your equation, y=x^2, notice that x^2 is always positive... so y must always be positive.

one for each result i got it sorry

gotcha

nice problem... thank you

aw no thank you