Below are two different functions, f(x)and g(x). What can be determined about their y-intercepts?

- Janu16

Below are two different functions, f(x)and g(x). What can be determined about their y-intercepts?

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- Janu16

f(x) = x + 4 x g(x)
-1 8
1 0
2 -4
@DanJS here it is

- DanJS

k

- DanJS

the y intercept is the point (x,y) = (0,y) when x is zero it will cross the Y axis, can you picture that

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- Janu16

um nope

- DanJS

|dw:1437444302167:dw|

- Janu16

oh i see

- DanJS

When x axis line is at zero, you are in the y-axis

- DanJS

so what can you think about f(x) for it's y-intercept, what would you do ?

- Janu16

so y is 4 right?..

- DanJS

right, you set X=0, then you get the y-intercept F(x) = 0 + 4

- Janu16

so put dot on 4?

- DanJS

yes you can place a point at ( x , y) = (0 , 4)

- DanJS

similarly, if you want to graph a line, you need 2 points right?
so find now the X-intercept by doing the same thing, this time set Y=0 and figure X

- Janu16

are you talking about g(x)?than it would be 1...?

- DanJS

f(x) = y = x + 4
0 = x + 4
when y is zero, x is -4
second point (x , y) = ( -4 , 0)

- Janu16

oh i see

- DanJS

no still f(x), similar question, now you can graph the line, you have 2 points, the X and the Y intercepts

- DanJS

|dw:1437444769726:dw|

- Janu16

why would the 4 be negative? for o,-4

- DanJS

fixed sorry

- Janu16

ohok

- DanJS

FOr g(x), it looks like it is also a line, constant slope

- Janu16

so the other 4 would be negative?i thought it would be positive

- Janu16

wasnt the point 4,0?

- DanJS

No, when y = f(x) = 0 , you get
0 = x + 4
0 - 4 = x + 4 - 4
x = -4
the x intercept is -4

- DanJS

plus the slope is +1, the line must be climbing

- Janu16

ohok i get it and what were you saying about g(x)?

- DanJS

if you calculate the slope between points 1-2 and 2-3 or any 2 of those points, the slope is the same, it is a line of the form y = m*x+ b

- DanJS

a line has the same slope everywhere

- DanJS

yep, now you hav e
y = mx + b
y = -4*x + b
pick any point and put in (x,y) and solve b

- DanJS

b is the y intercept for that form of a line equation, y = (slope)*x + (y int)

- DanJS

see when x = 0, you get y=b, the y intercept

- Janu16

so what can be determined about both of their y intercepts

- DanJS

since you figured the slope is -4 for g(x)
y = -4*x+ b
use any point on g, i pick (x , y ) = (1 , 0) ... put into equation
0 = -4*(1) + b
what is b - (y intercept) for g?

- Janu16

-4

- DanJS

|dw:1437445491201:dw|

- DanJS

b = 4
thjey both have the same y-intercepts

- DanJS

0 = -4 + b
add 4 both side
b = 4

- Janu16

ok they all have same y intercepts?

- DanJS

yes, if you don't get something let me know

- Janu16

i get it so my answer would be they all have same y intercept. thanks and you know how you helped me with f(g((x) problem i am trying to do mine but cant get the answer. f(x) = 2x + 1 and g(x) = −5x + 2, solve for f(g(x)) when x = 3. dont you start with g(x) first?

- Janu16

i did -5 times 3 + 2

- Janu16

nvm i got it its -25

- DanJS

yeah, remember to figure the inner function first, then use that value to move outwards

- Janu16

yup i remember and id you have g(f((x) do you start with f first?

- DanJS

yep.
W[f(G(x))] G then f then W, keeping track of the values to use for each as you work outwards

- Janu16

what is w

- DanJS

i just added a third function, same idea, work middle to outside

- Janu16

ohoki am just learning about 2 function right now in algebra 2 lol but thxs

- DanJS

no prob

- Janu16

does this \ mean graph is going negative? like that line

- DanJS

that means it has a negative slope, for increasing values of X, the Y value will be less and less

- DanJS

as you figured the slope was minus 4

- Janu16

ohok and i was doing inverse so how do you find inverse of this f(x) = x2 − 16 like it has only 2 things so how do you inverse that

- DanJS

If you see an equation in the form y = mx+b and the m is negative, it will do that

- Janu16

ok i see it and would it be x=y2-16 than what do you subtract

- DanJS

remember how the inverse reflects over that line with slope 1, y = x

- DanJS

##### 1 Attachment

- DanJS

yeah just solve for the new y value for the inverse if you need that

- Janu16

yas i see how it reflects but how do you solve for y value like do you subtracvt 16

- Janu16

i mean add

- DanJS

oh, you gots to get the algebra stuff down,, lol

- DanJS

\[x = y^2 - 16\]
\[x+16=y^2\]

- DanJS

take square root of both sides

- Janu16

±square root x+16?

- DanJS

yes that is y or f inverse

- DanJS

the + is above the x axis and the - is below, as graphed

- Janu16

so that would be the answer right?

- DanJS

right

- Janu16

thx :p

- DanJS

not sure what you have to remember for tests, but just recall inverse functions, switch x and y and solve new y is the inverse

- DanJS

if it exists, but it probably will if they ask you

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