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    (Original post by d1ck)
    I'd probably have to go back over my notes to try and figure this one out since I'm quite bad at translations Q5 part ii) - I would copy-paste the question but it won't let me :/
    f(x) = \sqrt{x} \Rightarrow \sqrt{x-4} =  f(x-4) which is a translation... to the right/left? How many units?


    also how would you q10 ii)

    I thought maybe I could put P's X coord into the circle equation to get a Y where they intersect, and P's Y coord to get an X. And then use y-y1 = m x-x1, but I don't have a gradient and tbh what I'm doing doesn't really make alot of sense to me :/
    You have two points: (5, -2) and (7, 2) - find the gradient of the line going through these two points then y - y_1 = m(x-x_1) finished it off. You do know how to find the line going through two given points, yeah? Or how to find the gradient of the line given two coordinates on it? Change in y over change in x.
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    (Original post by Riya06)
    Or you could just square the whole equation and solve it like a normal quadratic equation.

    Edit: This is wrong
    (Original post by Riya06)
    Oh? That's weird. You can't do that? :/
    You can do something similar: rearrange to get \sqrt{x} on one side then square and solve.
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    (Original post by Zacken)
    f(x) = \sqrt{x} \Rightarrow \sqrt{x-4} =  f(x-4) which is a translation... to the right/left? How many units?


    You have two points: (5, -2) and (7, 2) - find the gradient of the line going through these two points then y - y_1 = m(x-x_1) finished it off. You do know how to find the line going through two given points, yeah? Or how to find the gradient of the line given two coordinates on it? Change in y over change in x.
    to the right 4 units(?) btw how come you can just remove the square root sign in the f(x) bit?
    Spoiler:
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    i really hate translations brb crying

    yeh I know how to find the gradient lol, the only thing is when I put eg. x=7 into the circles equation it will come out with a quadratic right? can I square root the whole equation so it's (7-5) + (y+2) = 5 instead or would that make it wrong?
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    (Original post by d1ck)
    to the right 4 units(?) btw how come you can just remove the square root sign in the f(x) bit?
    Spoiler:
    Show
    i really hate translations brb crying
    yeh I know how to find the gradient lol, the only thing is when I put eg. x=7 into the circles equation it will come out with a quadratic right? can I square root the whole equation so it's (7-5) + (y+2) = 5 instead or would that make it wrong?
    Why are you putting x=7 into the equation of the circle?
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    (Original post by Zacken)
    Why are you putting x=7 into the equation of the circle?
    idek lol

    I thought that would give me coords to use y-y1 = m ( x-x1)
    do i just use the coords of P for that then?
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    (Original post by d1ck)
    do i just use the coords of P for that then?
    Yes.
 
 
 
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