Question

can someone help me with this

Answer 1

It’s been a while since I did this negl

Answer 2

Its college precalc ive been stumped on how to do this for awhile 😭

Answer 3

the formula for exponential is \[y=ab^x\]

Answer 4

a is the initial or y intercept

Answer 5

Yea I’m sorry man I don’t remember any of this i graduated last year

Answer 6

thanks anyways man

Answer 7

I just wanted to know how to do it Ill just get the answers then come back and learn how to do it some other time its due in 50 minutes

Answer 8

and i got like 2 questions left

Answer 9

I can do my best

Answer 10

I got it

Answer 11

It’s y=25/3

Answer 12

Wait no

Answer 13

Y=25 (3/5)

Answer 14

That’s your final answer

Answer 15

Okay this problem is messy, I can see why you'd have trouble:

You'd start off by plugging a given point into the equation, so we'll use (-1, 3/5)

\(\large \frac{3}{5}=ab^{-1} \)

From here, solve for "a"

\(\large \frac{3}{5} *b=a \)

Plug "a" into the second equation, which is obtained by plugging in the second coordinate points, (2, 25/9)

\(\large \frac{25}{9}=(\color{red}{\frac{3}{5}b} )b^3\)

Now combine and solve for "b"

\(\large \frac{25}{9}=\frac{3}{5}b^4 \)

\(\large \frac{25}{9}*\frac{5}{3}=b^4 \)

\(\large \frac{125}{27}=b^4\)

\(\huge \sqrt[4]{\frac{125}{27} }=b\)

Answer 16

Now the fun part is solving for "a," which we start by plugging back in "b" into the equation:

\(\huge \frac{25}{9}=a*(\sqrt[4]{\frac{125}{27}})^2 \)

The exponent and 4th root can be simplified

\(\huge \frac{25}{9}=a*(\sqrt{\frac{125}{27}}) \)

Now just divide and you get:

\(\huge \frac{ \frac{25}{9}} {( \sqrt{\frac{125}{27}}) } =a\)

Answer 17

Putting it together, you get this monstrosity as your answer:

\(\huge y= \frac{ \frac{25}{9}} {( \sqrt{\frac{125}{27}}) } * (\sqrt[4]{\frac{125}{27}})^x\)

I graphed it and the points at x=-2 and x=1 check out, but you might want to graph it too just to be sure.