the origin is an x-intercept if f(0) = 0.
Is that true here?
Suppose I'm graphing a graph like j(x) = -2 sin 4x. How can I tell if the origin is an x-intercept for it?
3 answers
Yes, the origin is an x-intercept here.
So how can I find the y-intercept in a function such as y = -2 sin(x - (pi/2)). I want to know from where to begin graphing, from the minimum or maximum value.
So how can I find the y-intercept in a function such as y = -2 sin(x - (pi/2)). I want to know from where to begin graphing, from the minimum or maximum value.
hey, come on. The y-intercept is where x=0!
So, just plug it in
y(0) = -2 sin(0-pi/2) = -2(-1) = 2
getting x-intercepts can be hard. Getting the y-intercept is trivial! Don't forget your Algebra I now that you're in trig.
So, just plug it in
y(0) = -2 sin(0-pi/2) = -2(-1) = 2
getting x-intercepts can be hard. Getting the y-intercept is trivial! Don't forget your Algebra I now that you're in trig.
1 answer