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Mathematics 17 Online
OpenStudy (anonymous):

Write a formula for the distance between a point (x, y) on the graph of y = x^4 and the point (3, 6). Express your answer in terms of x.

OpenStudy (anonymous):

distance between points (3, 6) an (x, x^4) is done with the distance formula: Distance = \(\large \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \)

OpenStudy (anonymous):

so plug in your x coordinates of 3 and x and your y coordinates of 6 and x^4 into the formula...

OpenStudy (anonymous):

But how would you get anything from the x^4? Thats what I'm confused about.

OpenStudy (anonymous):

the point (x, x^4) is on the graph of y = x^4... didn't you say you wanted an expression (in terms of x) for the distance between (3, 6) and a point on the graph?

OpenStudy (anonymous):

yes.

OpenStudy (anonymous):

What do I do after I plugged the points in the distance formula? How do I get the answer?

OpenStudy (anonymous):

I have square root of (3-x)^2 + (6-x^4)^2

OpenStudy (anonymous):

that's the expression for the distance between (3, 6) and a point on the graph for whatever value x has....

OpenStudy (anonymous):

for example.. when x=2, the point (2, 2^4) or (2, 16) is on the graph of y=x^4.... so the distance between (3, 6) and (2, 16) is: D = \(\large \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}=\sqrt{(3-2)^2+(6-16)^2} \)

OpenStudy (anonymous):

So you are saying that that's my expresstion is the distance formula?

OpenStudy (anonymous):

basically... yes....

OpenStudy (anonymous):

but make it in terms of only x....

OpenStudy (anonymous):

yes I got it. Thank you

OpenStudy (anonymous):

yw...:)

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