Question

[I'm having trouble with part b. how is this part set up on my calculator?
Let p=(x,y) be a point on the graph of y=v¯(x)
a) Express the distance d to P to the point (1,0) as a function of x
My answer: v¯(x^2-x+1)
b) Use a graphing utility to graph d=d(x)

Answer 1

do you know what
\[v(x)\] is?

Answer 2

no

Answer 3

what ever it is, the distance will be
\[\sqrt{(x-1)^2+v^2(x)}\]
and it will be a lot easier to graph and work with if you graph the square of the distance instead, i.e. graph
\[(x-1)^2+v^2(x)\]

Answer 4

if you do not have an expression for
\[v(x)\] then there is no way to graph this.

Answer 5

oh sorry v¯(x^2-x+1) is square root of (x^2-x+1) not v(x)

Answer 6

i was trying to make a square root sign

Answer 7

ooooooooooooooooooh i see hold on we can work this out

Answer 8

the distance between
\[(1,0)\] and
\[(x,\sqrt{x^2-x+1})\] right?

Answer 9

messed that up. we get
\[d^2=(x-1)^2+x^2-x+1\]
\[d^2=x^2-2x+1+x^2-x+1\]
\[d^2=2x^2-3x+2\] that looks better

Answer 10

and as i mentioned it would be easier to work with the square of the distance than the distance, but if you actually need the distance itself, use
\[d=\sqrt{2x^2-3x+2}\]

Answer 11

k