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)#

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#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)#

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#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#