At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
is this a calculus question?
if c=1/2 then there is only one line... that is the y-axis. if c<1/2, there are no lines because the parabola will always be positive. Now if c>1/2, there are always two lines.
how do you figure
take the derivative of y=x^2 to get y'=2x. This gives the slope of the tangent line for what ever x coordinate you look at that's on the parabola. Since you want the line that's normal to the parabola, take the negative reciprocal of the derivative to get you the slope of the line perpendicular to the tangent line. Since you know the line goes through (0, c) you can use that as the point to get the equation of the line perpendicular to tangent line using point-slope form of a line. you should end up with this equation: y = c - 1/2. this is how I came to the conclusions in my previous post.
so the slope of the normal line is -1/(2x)??
i thought the slope had to be a number
like so that's equal to a slope of -1/2?
yes but the slope on a parabola depends on what point you look at. so the slope is given as an expression.
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
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
so if c = 1/2, y = 0, and... how does that mean that there is one line and that it's the y-axis?
when y=0, this makes x=0. so the line normal to the parabola at (0,0) is the vertical line, x=0, and this is the only one when c=1/2
oh.. k. so if c < 1/2, it's negative, and x^2 is always positive on y... how would i check that otherwise though if i didn't know the graph of the function?
from your equation, y=x^2, notice that x^2 is always positive... so y must always be positive.
but uh i'm sorry i know i must seem stupid as hell but how does that (c = 1/2, y = 0, x = 0, etc) mean that x is always 0 on the line? that's not just a point? geez....
that i know i was asking like if i didn't know the graph of the function like i know the graph of x^2... but obviously it's no big deal, same deal, it's gotta fit in the domain/range..... but uh there i'm a little fuzzy anyway what the hell is c in the normal line
it does not mean that x is always zero, all I did was to solve y=x^2 when y=0. Now, if c>1/2, notice that your y is positive so that yields two x values, one positive and the other negative.
alright but so what is that you're finding? how is the resulting x,y a LINE and not a POINT? you know what i mean? that's how i'm confused
oh they're each other points on the normal line that goes through the point (0,c) so that makes a line
one for each result i got it sorry
remember the question was "how many lines..." ? the resulting points gave you the other point(s) needed to connect with (0, c) and be normal to the parabola...
nice problem... thank you
aw no thank you