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Mathematics 13 Online
OpenStudy (anonymous):

Write a formula for the distance between a point (x, y) on the graph of y = x4 and the point (4, 8). Express your answer in terms of x

OpenStudy (anonymous):

The shortest distance between a point on this graph and a point somewhere else, will occur at a perpendicular to the graph. Can you find the slope (as a function) of your function?

OpenStudy (anonymous):

slope is (8-y)/(4-x) right?

OpenStudy (anonymous):

That is a slope... are you in Calculus?

OpenStudy (anonymous):

yes thats what im asking you

OpenStudy (anonymous):

2nd day in calc

OpenStudy (anonymous):

Awesome... you can also find a slope with a derivative. Ahhh, second day. So this one is gonna be rough.

OpenStudy (anonymous):

ok so if i have the slope what do i do next?

OpenStudy (anonymous):

I made an equation for a line to start... but that is not going to help you. I would say that you would use a distance formula.\[\sqrt{(y _{2}-y _{1})^{2}+(x _{2}-x _{1})^{2}}\]

OpenStudy (anonymous):

y2 would be your function, x2 would just be x. (x1,y1) is your point

OpenStudy (anonymous):

should i just make a distance formula for just point x,y and 4,8? but what about x^4

OpenStudy (anonymous):

\[d(x)=\sqrt{(x ^{4}-8)^{2}+(x-4)^{2}}\]

OpenStudy (anonymous):

This way, the first point is (4, 8) and the second point is\[(x, x ^{4})\]

OpenStudy (anonymous):

oh i got it

OpenStudy (anonymous):

thank you so much babe

OpenStudy (anonymous):

I thought that it was a different question at first... sorry for the lag. I had to get my bearings and start from the beginning. You're welcome.

OpenStudy (anonymous):

i know i was confuse because i was trying to find the slope and do a mx+b equation

OpenStudy (anonymous):

Lol! Yup that's what I did... der.

OpenStudy (anonymous):

can you help me with a new question? :)

OpenStudy (anonymous):

ok

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