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# P2 Jan 04 watch

1. questionn 4):

the function f is even and has domain R. For x is greater or equal to 0, f(x) = x^2 - 4ax, where a is a +ve constant.

a) sketch the curve showing all coordinates at which curve meets the axes.

i never am able to sketch things with a constants because you obviously cant plot it. can neone help me on how i go about sketching with unknown constants in functions?
2. The easiest way for this is to complete the square so f(x) = (x - 2a)^2 - 4a^2 so the curve is a parabola shifted 2a to the right and 4a^2 down

To find the intercepts, first sub in x=0, f(0) = 0 so y-intercept is 0. Then factorise f(x) = x(x - 4a) so curve crosses the x-axis at x = 0 and x = 4a.
3. (Original post by Jubba)
...

i never am able to sketch things with a constants because you obviously cant plot it. can neone help me on how i go about sketching with unknown constants in functions?
You can plot it by letting the constant a=1 or any other "easy" value, that looks keeping the function simple-looking and then see what that looks like.
4. (Original post by Jubba)
questionn 4):

the function f is even and has domain R. For x is greater or equal to 0, f(x) = x^2 - 4ax, where a is a +ve constant.

a) sketch the curve showing all coordinates at which curve meets the axes.

i never am able to sketch things with a constants because you obviously cant plot it. can neone help me on how i go about sketching with unknown constants in functions?
this is a stupid question
5. (Original post by Jubba)
questionn 4):

the function f is even and has domain R. For x is greater or equal to 0, f(x) = x^2 - 4ax, where a is a +ve constant.

a) sketch the curve showing all coordinates at which curve meets the axes.

i never am able to sketch things with a constants because you obviously cant plot it. can neone help me on how i go about sketching with unknown constants in functions?
well it is x^2 parabola, and it is shifted down and right (but still passing through 0) so that the x-intercept is 4a..

Think about it, when x = 0, y = 0..
Factorise:

x^2 - 4ax = x(x - 4a)
x can equal 4a and give f(x) = 0 then, so the place where it crosses the x axis a second time is at 4a.

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