Explanation:
you could answer the main complex question correctly by figuring out the resulting function g(x), but you don't understand the rather simple additional questions about some minor details ?
are you serious ?
so, what you have figured out :
g(x) = -f(x + 3) - 5
the reflection across the x-axis out the sign of the functionaries results upside-down. therefore, the "-" sign in front of f.
the downward shift by 5 units added the "- 5" at the end, because all functional result values are lowered by 5 units (that's what the shift downward by 5 units means).
and the shift left by 3 units made us transform x to x + 3 as the input value (because everything happens now 3 units earlier than for the original f(x)).
what is missing now is to use the original f(x) = 3^x
g(x) = -f(x+3) - 5 = -(3^(x+3)) - 5 = -(3^x)×3³ - 5 =
= -27×(3^x) - 5
either one of the last 3 expressions is the final g(x). pick the one you think your teacher is looking for the most.
the y-intercept of a curve/function is the functional value when x = 0 (that is what it intercepts the y-axis).
so,
y-intercept = g(0) = -27×(3^0) - 5 = -27×1 - 5 = -27 - 5 = -32
the domain of the function is the interval or set of all valid values for x.
do we need to exclude any negative or positive values that cannot be calculated in the expression, as it lead to an undefined value or situation ? 0 ?
no, any number we can think of is fine for x in the functional expression.
so, the domain is
x is in (-infinity, +infinity).
please note the round brackets, as the interval ends are not included. simply because infinity is not a defined number.
the range of a function is the interval or set of all build values for y, the functional results.
can the function ever deliver a positive value ?
no. -27×(3^x) - 5 will always be negative, no matter what number (positive it negative) we use for x.
can the function ever be 0 ?
no. only for x = -infinity. and remember, this is not a valid number. x will never reach -infinity (only in the infinity future or past), and therefore the functional value can get closer and closer to 0 bu will never reach it.
so, the range is
y in (-infinity, 0).
again with the round brackets, because the interval limits will never be included.