um i need help my math book doesnt explain this
Graph
2007-10-29 21:54:30 · 7 answers · asked by theguywiththebigstick- What now! 3 in Science & Mathematics ➔ Mathematics
it looks like this:
http://id.mind.net/~zona/mmts/functionInstitute/rationalFunctions/oneOverX/oneOverX.html
2007-10-29 22:05:20 · answer #1 · answered by Theta40 7 · 0⤊ 0⤋
I don't understand what you want to do. Do you just want to graph it? If so, you can just make it y = 4/x and plug in some values for x to get some points and plot the graph. For example y(1) = 4/1 = 4...so one point is going to be (1,4). It's also important to note that x cannot equal 0 (or you'd get 4/0 which is undefined). So the domain of the function is all real numbers SUCH THAT x does not equal 0.
If you want to sketch it there are various techniques depending on what level of math you are at.
2007-10-30 05:01:34 · answer #2 · answered by JoeSchmo5819 4 · 1⤊ 0⤋
the graph of 1/x if a rectangular hyperbola in the 1st and 3rd quadrant.
it will be asymptoting (i.e approaching towards the axes but never touching them) in both the quadrants towards infinity.
now if we multiply it with 4, we have
f(x)=4/x.
this graph can be obtained by multiplying every point in the above graph with 4.
2007-10-30 05:01:50 · answer #3 · answered by crashbird 2 · 1⤊ 0⤋
its a function of linear equations. it can be read that the function of x is 4/x. if you are given a value of x just substitute it to all the x's in the function. for example, if x=2 then f(2)=4/2
so if you wud evaluate: f(2)= 2
2007-10-30 05:03:10 · answer #4 · answered by Anonymous · 0⤊ 0⤋
I don't get it. It's not a quadratic equation. A quadratic equation has an x^2 coefficient
2007-10-30 04:58:12 · answer #5 · answered by terrorblade 3 · 0⤊ 0⤋
a function passes through (1,4) (2,2), (4,1)
y=0 asymptote
a curve line looks up
2007-10-30 04:59:52 · answer #6 · answered by iyiogrenci 6 · 1⤊ 0⤋
I think that's a hyperbola.
2007-10-30 05:15:42 · answer #7 · answered by Shh! Be vewy, vewy quiet 6 · 0⤊ 0⤋