Question

Finding standard function?

Answer 1

1) y=f3(x)=e^x-2
2)y=f4(x)=ln(x+2)
3)y=f5(x)=-x^3+1

Answer 2

@jim_thompson5910

Answer 3

the first one is this right?

\[\Large y = f_3(x) = e^{x-2}\]

or is it this?

\[\Large y = f_3(x) = e^{x}-2\]

Answer 4

the second! \[(e^x)-2\]

Answer 5

ok and what instructions did the professor post for this one? s/he said to graph this? and to find the standard form?

Answer 6

just graph and identify a point, i figured finding standard form was part of the proccess

Answer 7

oh no need for standard form if all you need is a point or two

to find any point, you first plug in any x value you want. This x value must be in the domain of f(x)

the good news is that the domain of (e^x) - 2 is the set of all real numbers

Answer 8

an easy x value is 0


y = (e^x) - 2

y = (e^0) - 2 ... replace every x with 0

y = 1 - 2 ... x^0 = 1 where x is nonzero

y = -1


So when x = 0, the value of y is y = -1
Therefore, we know that the point (0,-1) is on the graph of y = (e^x) - 2

Answer 9

does that make sense?

Answer 10

Yes it does!
so for the second and 3 i could just go ahead and plug any value for x and plot

Answer 11

which x value do you want to plug in?

Answer 12

i can do x=-1 with y coming out to 0 and then another that would make a real #

Answer 13

x=53 for example gives y=4

Answer 14

\[\Large y = e^x - 2\]
\[\Large y = e^{-1} - 2\]
\[\Large y = ???\]
the answer won't be y = 0

Answer 15

oh sorry i have moved on to the next one!!

Answer 16

for that one i did (-28,2) and (0,-1)

Answer 17

I see

Answer 18

so for ln(x+2) you plugged in x = 53 ?

Answer 19

yes and got 4.00...

Answer 20

y = 4.007, good

Answer 21

What the graphing calculator (like desmos) is doing is computing a whole bunch of (x,y) pairs. Hundreds possibly thousands. Then it plots them all on the same grid. After that, it connects the dots. The more points, the more accurate the graph.