# Use the function h(x)=3(.83^3x) for the following problems: 1.) ___ This is (a) an exponential growth function (b) a logarithmic function (c) an exponential decay function (d) a function using...

Use the function h(x)=3(.83^3x) for the following problems:

1.) ___ This is (a) an exponential growth function (b) a logarithmic function (c) an exponential decay function (d) a function using natural logarithms

2.) ___ The y-intercept (a) is at y≈.57 (b) is at y=.83 (c) is at x=3 (d) does not exist (e) none of the above

3.) ___ The x-intercept (a) is at x=0 (b) is at x≈.57 (c) is at y=3 (d) does not exist (e) none of the above

4.) The end behavior of this function is best described as:

(a) h(x)-> ∞ as x->- ∞; h(x)->0 as x->  ∞

(b) h(x)->0 as x->- ∞; h(x)-> ∞ as x-> ∞

(c) h(x)-> - ∞ as x->- ∞; h(x)->0 as x->  ∞

(d) h(x)->0 as x->- ∞; h(x)->- ∞ as x-> ∞

For #5 use the following equation: k(x)=(x-2)^2+2

5.) Choose all the true statements:

(a) k(x)>h(x) for x≤0

(b) as x-> ∞, k(x) > h(x)

(c) as x->-∞, h(x) > k(x)

(d) k(x)>h(x) for |x|≤1

(e) h(x)>k(x) for -1 <x≤5

(f) as x-> 2, h(x)>k(x)

ishpiro | Certified Educator

1) This is an exponential function

2) Y-intercept is determined by y-value when x = 0. So, `y = h(0) = 3*0.83^(3*0)=3`

The y-intercept of h(x) is y = 3, but there is no answer choice like this. So it is either x = 3 which was meant to be typed "y = 3", or "none of the above".

3) X-intercept does not exist, because the graph of h(x) never intersects x-axis. h(x) is not 0 for any x.

4) The end behavior: when x is approaching positive infinity, that is, becoming very large, h(x) is also becoming very large and approaches infinity. When x is approaching negative infinity, h(x) is becoming very small and approaches 0, but never reaches it. So the answer is b).

ishpiro | Certified Educator

4) Correction: the answer is a). Since the base of exponent is less than 1, h(x) approaches 0 when x is very large (approaches positive infinity) and h(x) approaches positive infinity when x is large and negative (approaches negative infinity.)