# Thread: [SOLVED] P=(x,y) on the graph of y=sqrt(x)...keep coming up with wrong answer

1. ## [SOLVED] P=(x,y) on the graph of y=sqrt(x)...keep coming up with wrong answer

Let P=(x,y) be a point on the graph of y=sqrt(x). Express the distance d from P to the point (1,0) as a funtion of x.

so I did:
f(x)=sqrt((x^2)+(y^2))
f(x)=sqrt((x^2)+(sqrt(x)^2))
f(x)=sqrt(x^2+x)
f(x)=x(x+1)

a) d(x) = x^2 +2x+2
b) d(x) = sqrt(x^2-x+1)
c) d(x) = sqrt(x^2-2x+2)
d) d(x) = x^2 -x+1

What am I doing wrong?

2. The distance from any point $(a,b)$ to the point $(1,0)$ is given by $\sqrt{(1-a)^2 + (0-b)^2}$.

3. Originally Posted by ninobrn99
What am I doing wrong?
You're finding the distance from the origin, not from $P$.

You should have

$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$= \sqrt{(x - 1)^2+(y-0)^2}$

$= \sqrt{(x-1)^2+\left(\sqrt x\right)^2}$

4. Originally Posted by ninobrn99
Let P=(x,y) be a point on the graph of y=sqrt(x). Express the distance d from P to the point (1,0) as a funtion of x.

so I did:
f(x)=sqrt((x^2)+(y^2))
f(x)=sqrt((x^2)+(sqrt(x)^2))
f(x)=sqrt(x^2+x)
f(x)=x(x+1)

a) d(x) = x^2 +2x+2
b) d(x) = sqrt(x^2-x+1)
c) d(x) = sqrt(x^2-2x+2)
d) d(x) = x^2 -x+1

What am I doing wrong?
Okay, why did you do that? What happened to the point (1, 0)? What you give (until your last line) is the distance from $(x, \sqrt{x})$ to (0, 0). But your last line is wrong- $x^2+ x= x(x+1)$, not $\sqrt{x^2+ x}$. For that you still need the square root: $\sqrt{x^2+ x}= \sqrt{x(x+1)}$.

But as I said, that would give the distance to (0, 0), not (1, 0).
Start with $d= \sqrt{(x- 1)^2+ (y- 0)^2}= \sqrt{(x- 1)^2+ y^2}$ and then replace $y^2$ with x.

5. Originally Posted by Plato
The distance from any point $(a,b)$ to the point $(1,0)$ is given by $\sqrt{(1-a)^2 + (0-b)^2}$.
im coming up with:
d(x) = sqrt((a^2-2a+1) + b^2)

6. Originally Posted by ninobrn99
im coming up with:
d(x) = sqrt((a^2-2a+1) + b^2)
Yes, and in this case $(a,\,b)=\left(x,\,\sqrt x\right),$ so substitute and simplify.

7. Originally Posted by Reckoner
You're finding the distance from the origin, not from $P$.

You should have

$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$= \sqrt{(x - 1)^2+(y-0)^2}$

$= \sqrt{(x-1)^2+\left(\sqrt x\right)^2}$
ok, so then :
$=\sqrt{x^2-x+1}$

Originally Posted by HallsofIvy
Okay, [b]why[b] did you do that? What happened to the point (1, 0)? What you give (until your last line) is the distance from $(x, \sqrt{x})$ to (0, 0). But your last line is wrong- $x^2+ x= x(x+1)$, not $\sqrt{x^2+ x}$. For that you still need the square root: $\sqrt{x^2+ x}= \sqrt{x(x+1)}$.

But as I said, that would give the distance to (0, 0), not (1, 0).
Start with d= \sqrt{(x- 1)^2+ (y- 0)^2}= \sqrt{(x- 1)^2+ y^2}[/tex] and then replace $y^2$ with x.
I did what I did because I was following the example in the book. That's what was throwing me off. This book doesn't explain thouroughly enough. It just shows a few examples and that's it. I usually use mathway, but I am really struggling with these last 3 word problems!!