Given: #1.1(3)^x#
Just because I prefer it this way write as # y=11/10xx3^x#
#color(blue)("Determine the "y_("intercept"))#
The # y_("intercept") # occurs at #x=0#
However, #3^0=1# giving
#color(brown)(y_("intercept")=11/10xx1 = 11/10)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider the case "x<0)#
This changes #3^x# into some root of #3#
So as #x# becomes increasingly negative #3^x# becomes less and less. So as #x# tends to negative infinity then:
#color(brown)(y=11/10xxlim_(x->oo^(-))(3^x)color(white)("ddd") ->color(white)("ddd")y=11/10xx0=0)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Consider the case "x>0)#
As #x# becomes increasingly greater the #3^x# increases exponentially.
#color(brown)(y=11/10xxlim_(x->oo^(+))(3^x)color(white)("ddd") ->color(white)("ddd") y=11/10xxoo = oo#
