Graph the function and analyze it for domain, range, continuity, increasing or decreaseing behavior, symmetry, boundedness, extr
ema, asymptotes, and end behavior.
f(x) = 3 * 0.2^x
f(x) = 3 * 0.2^x
1 answer:
Answer:
Step-by-step explanation:
Domain of f: "What values of x can we plug into this equation?" This makes sense for all real numbers so the domain is
Range of f: "What values of f(x) can we get out of the function?" From the graph we see we can get any real number greater than 0 out of the function by choosing a suitable x-value in the domain. The range is therefore .
Continuity: Since the graph is one, unbroken curve (i.e. a curve that can be drawn in one movement without taking your pen off the paper). We see that "roughly speaking" the function is continuous.
Increasing or decreasing behaviour: For all x in the domain, as x increases, f(x) decreases. This means the function exhibits decreasing behaviour.
Symmetry: It is clear to see the graph of f(x) has no symmetry.
Boundedness: Looking at the graph we see it is unbounded above as when we choose negative values, the graph of f(x) explodes upwards exponentially. Choose a value of x, plug it in, next choose (x-1), plug this in and we observe for all x in the domain.
The function is however bounded below by 0: no value of x in the domain exists which satisfies .
Extrema: As far as I can tell, there are no turning points on the curve. (Is this what you mean by extrema?)
Asymptotes: Contrary to the curve's appearance, there are no vertical asymtotes for this curve. The negative-x portion of the curve is just growing so quickly it appears to look like an asymptote. There is a value of f(x) for all x<0. There is however a horizontal asymtote: .
End behaviour: As . As
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I'm going to call the function for the graph of A, A(x) and the function for the graph of B, B(x),and the function for the graph of C, C(x)
we know that when A(x)=0 (and x>0), C(x)=B(x) at that value of x
also, when B(x)=0, A(x)=C(x) at that value of x
we need to find the intersection points
a quadratic function can be written in form y=a(x-b)(x-c) where the roots or zeroes are at x=b and x=c and a is a constant. since we can only find 2 points, we will assume a=1 since in the other quadratic function, a=1 as well
let's use our 1st discovery first
when A(x)=0, (and x>0), C(x)=B(x)
find zeroes of A(x)
find where it equals 0
, factor, what 2 numbers mulitly to get -3 and add to get 2
0=(x-1)(x+3)
set equal to 0
0=x-1, x=1
0=x+3, x=-3
we are looking for positive root, so at x=1, C(x)=B(x)
meaining C(1)=B(1), 3(1)+3=B(1), 3+3=B(1), B(1)=6, that's one point
2nd thing
when B(x)=0, A(x)=C(x)
solving A(x)=C(x) for x<0 (since it's to the left of y axis)
factor, what 2 numbers multiply to get -6 and add to get -1
(x-3)(x+2)=0
set equal to 0
x-3=0, x=3
x+2=0, x=-2
we want x<0 so at x=-2
find y value, C(3)=3(-2)+3=-6+3=-3
we know that B(x) has a zero at (-3,0)
we also know that B(x) passes through the point (1,6)
if a quadratic function has zeroes at m and n, then it can be factored into form y=a(x-m)(x-n) where a is a constant, we assume a=1 because otherwise we can't find the function since we only know 2 points
we know that one zero is at -3,
y=(x-(-3))(x-n)
y=(x+3)(x-n)
subsitute the other point (1,6) to find n
6=(1+3)(1-n)
6=(4)(1-n)
6=4-4n
2=-4n
n=-1/2
so the equation is
or
the only problem is that in the picture, the graph of B crosses the x axis at a positive and negative value but the equation I have crosses the x axis at 2 negative values
but without further information, I can't solve for B