Question 1153574
SHIFTS OF THE PARENT FUNCTION:

{{{ f(x)=a*b^x}}}
​​ 
For any constants  {{{a}}}, {{{c}}} and {{{d}}}, the function 

{{{ f (x )= a*b ^(x+c)+d}}}

-is stretched vertically by a factor of {{{a}}} if {{{abs(a)>1}}} 
-is compressed vertically by a factor of {{{a}}} if {{{abs(a)<1}}} 
-has a y-intercept of   ({{{0}}},{{{a}}})
​​ -shifts the parent function vertically {{{d}}} units, in the same direction of the sign of {{{d }}}
-horizontally {{{c}}} units, in the opposite direction of the sign of {{{c}}} 
-the y-intercept becomes  ({{{0}}},{{{ b ^ (c +d)}}})
-the horizontal asymptote becomes {{{y = d}}}
-the range becomes ({{{d}}},{{{infinity}}}).
-the domain, ({{{-infinity}}} ,{{{infinity}}}), remains unchanged 


in your case we have {{{f(x) = -(1/2) (1/6)^( x-7) +9}}}


the equation of the new function, {{{g(x)}}}:

- is shifted left {{{7 units}}}, so we have

{{{g(x) = -(1/2) (1/6)^( x-7+7) +9}}}

{{{g(x) = -(1/2) (1/6)^x +9}}}


-stretched vertically by a factor of{{{ 6}}}, so we have

{{{g(x) = 6(-1/2) (1/6)^x +9}}}

{{{g(x) = -3*(1/6)^x +9}}}


-reflected about the {{{x}}}-axis, so we have

{{{g(x) =(-1)*( -3)*(1/6)^x +9}}}

{{{g(x) = 3*(1/6)^x +9}}}


and then shifted downward {{{9}}} units, so we have

{{{g(x) = 3*(1/6)^ x +9-9}}}

{{{g(x) = 3*(1/6)^x }}}



State the y-intercept of {{{g(x)}}}.

({{{x}}}, {{{y}}}) =({{{0}}},{{{3}}})


 the domain and range of {{{g(x)}}}:

the range becomes ({{{0}}},{{{infinity}}}) 
the domain, ({{{-infinity }}},{{{infinity}}})


{{{ graph( 600, 600, -10, 15, -10, 15, -(1/2) (1/6)^( x-7) +9,3*(1/6)^x) }}}