+50 Write a tangent function h(x), such that the halfway points on one period are (-pi/2 , 2) and (pi/2 , 4)
+118654 question: I still don't get it.
3+1=4
and
3-1=2
y=2 and y=4 are on either side of the centre - the middle is y=3
so (0,3) must be the 'centre' of your graph,
if the graph was y = tanx then one centre is (0,0)
If you want the centre to be (0,3) it will have to be y= (tanx) +3
does that help?
+118654 Write a tangent function h(x), such that the halfway points on one period are (-pi/2 , 2) and (pi/2 , 4)
Mmm not even sure what this means Ninja.... what is a halfway point and how can you have 2 of them?
Maybe
\(y=3+sinx\)
full graph
https://www.desmos.com/calculator/xn30b7wnd2
image only
+118654 Since it is tan maybe this is what is wanted. It still doesn't make much sense though.
y=3+tan(x/2)
https://www.desmos.com/calculator/0fh3otmzqj
+118654 Ninja's question is
"Where the heck did you get the 3 from" :D
y=tanx is centred on (0,0) (at least on of the periods is)
so if you wanted points rotationally symmetrical on either side they would be
(0-x,0-y) and (0+x,0+y)
so they could be
your points are (-pi/2 , 2) and (pi/2 , 4)
The y values are 2 and 4 so the centre of them is 3 Hence the graph has be lifted up 3 .
That is where the 3 came from :/
+118654 question: I still don't get it.
3+1=4
and
3-1=2
y=2 and y=4 are on either side of the centre - the middle is y=3
so (0,3) must be the 'centre' of your graph,
if the graph was y = tanx then one centre is (0,0)
If you want the centre to be (0,3) it will have to be y= (tanx) +3
does that help?
