What do each of these variables mean? (Exponential functions)

[supernova1203]

Homework Statement

I've been working with these problems for the past few days and although i managed to solve them all, I'm still not a 100% sure so i'd like someone to clarify for me

What do EACH of these variables mean? And on a graph what do they do?

a

b

d

c

k

in the equation y= ab^k(x-d)+c

Homework Equations

y=ab^k(x-d)+c

for example we were given a problem y=2-3(5^x+4)...here i was having trouble determining which variable was which...until i looked at the base function which was y=5^x and realized they are of the form y=b^x, and realized that 5 was the B value...and 2 was the C value in y=ab^k(x-d)+c they had just re arranged it to confuse me

or this one that gave me quite a bit of trouble...

y=1/5(32^x)-4..
How would one know which is which here?
i suppose -4 is the c value...and 32 is the k value? and 1/5 is the b value or a value? and the d value is 1 because there's nothing next to the x which implies a 1.

the 1/5 was the one that confused me the most i couldn't determine weather it was the a or the b value...and i'm not even sure what the b value does...



sorry if this is not the most coherent question

The Attempt at a Solution

the a value stretches the graph vertically, and if negative reflects on the x axis

the d value determines if the graph will shift on the x axis(left or right)

the c value determines if the graph will shift on the y axis(up or down)

the k value determines the stretches of the graph horizontally and if negative reflection on the y axis...


still got no clue what the b value does...

 

[chaoseverlasting]

Are you sure that 'c' is in the exponent power? From the looks of it, a seems to be the amplitude, k is the damping factor, c is a constant which just shifts the graph vertically and d is just a phase shift.

All this is with reference to signals. From pure mathematics, c remains unchanged, d just shifts the graph to the left or right, k controls how fast the graph rises, and a is the amplitude.

 

[supernova1203]

Are you sure that 'c' is in the exponent power? From the looks of it, a seems to be the amplitude, k is the damping factor, c is a constant which just shifts the graph vertically and d is just a phase shift.

All this is with reference to signals. From pure mathematics, c remains unchanged, d just shifts the graph to the left or right, k controls how fast the graph rises, and a is the amplitude.

k determines horizontal stretch, a determines vertical stretch, d and c determine where the graph will be moved to(Although they do not determine the shape of the graph like a and k do)
d determines where it will be moved on the x-axis and c determines where it will be moved on the y axis.

 

[eumyang]

y= ab^k(x-d)+c

You mean this, right?
y = a \cdot b^{k(x - d)} + c

y=1/5(32^x)-4..
How would one know which is which here?
i suppose -4 is the c value...and 32 is the k value? and 1/5 is the b value or a value? and the d value is 1 because there's nothing next to the x which implies a 1.

the 1/5 was the one that confused me the most i couldn't determine weather it was the a or the b value...and i'm not even sure what the b value does...

b, the base, is the value that is raised to x, or the expression that contains x. So b is not 1/5. a is the vertical stretch/shrink factor, as you said, and it's the number multiplied by the base. You didn't specify a value for a.

k is the number multiplied by x as part of the exponent, so k is not 32. d is the number added/subtracted from x as part of the exponent, so d is not 1.

This may be too big of a hint, but if I rewrite the function as
y = \frac{1}{5} \cdot 32^{1(x - 0)} + (-4)
can you find the correct values now?

 

[supernova1203]

You mean this, right?
y = a \cdot b^{k(x - d)} + c


b, the base, is the value that is raised to x, or the expression that contains x. So b is not 1/5. a is the vertical stretch/shrink factor, as you said, and it's the number multiplied by the base. You didn't specify a value for a.

k is the number multiplied by x as part of the exponent, so k is not 32. d is the number added/subtracted from x as part of the exponent, so d is not 1.

This may be too big of a hint, but if I rewrite the function as
y = \frac{1}{5} \cdot 32^{1(x - 0)} + (-4)
can you find the correct values now?

damnit you are right...also where did you learn tex? It's something i want to learn

 

[supernova1203]

<br /> $\begin{array}{lll}<br /> y = \frac{1}{5} \cdot 32^{1 (x - 0)} + 4 &amp; &amp; \grave{}<br /> \end{array}$<br />

omg... i did that! ^^ I am learning tex!