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the y intercept is the point (x,y) = (0,y) when x is zero it will cross the Y axis, can you picture that
oh i see
When x axis line is at zero, you are in the y-axis
so what can you think about f(x) for it's y-intercept, what would you do ?
so y is 4 right?..
right, you set X=0, then you get the y-intercept F(x) = 0 + 4
so put dot on 4?
yes you can place a point at ( x , y) = (0 , 4)
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
are you talking about g(x)?than it would be 1...?
f(x) = y = x + 4 0 = x + 4 when y is zero, x is -4 second point (x , y) = ( -4 , 0)
oh i see
no still f(x), similar question, now you can graph the line, you have 2 points, the X and the Y intercepts
why would the 4 be negative? for o,-4
FOr g(x), it looks like it is also a line, constant slope
so the other 4 would be negative?i thought it would be positive
wasnt the point 4,0?
No, when y = f(x) = 0 , you get 0 = x + 4 0 - 4 = x + 4 - 4 x = -4 the x intercept is -4
plus the slope is +1, the line must be climbing
ohok i get it and what were you saying about g(x)?
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
a line has the same slope everywhere
yep, now you hav e y = mx + b y = -4*x + b pick any point and put in (x,y) and solve b
b is the y intercept for that form of a line equation, y = (slope)*x + (y int)
see when x = 0, you get y=b, the y intercept
so what can be determined about both of their y intercepts
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?
b = 4 thjey both have the same y-intercepts
0 = -4 + b add 4 both side b = 4
ok they all have same y intercepts?
yes, if you don't get something let me know
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?
i did -5 times 3 + 2
nvm i got it its -25
yeah, remember to figure the inner function first, then use that value to move outwards
yup i remember and id you have g(f((x) do you start with f first?
yep. W[f(G(x))] G then f then W, keeping track of the values to use for each as you work outwards
what is w
i just added a third function, same idea, work middle to outside
ohoki am just learning about 2 function right now in algebra 2 lol but thxs
does this \ mean graph is going negative? like that line
that means it has a negative slope, for increasing values of X, the Y value will be less and less
as you figured the slope was minus 4
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
If you see an equation in the form y = mx+b and the m is negative, it will do that
ok i see it and would it be x=y2-16 than what do you subtract
remember how the inverse reflects over that line with slope 1, y = x
yeah just solve for the new y value for the inverse if you need that
yas i see how it reflects but how do you solve for y value like do you subtracvt 16
i mean add
oh, you gots to get the algebra stuff down,, lol
\[x = y^2 - 16\] \[x+16=y^2\]
take square root of both sides
±square root x+16?
yes that is y or f inverse
the + is above the x axis and the - is below, as graphed
so that would be the answer right?
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
if it exists, but it probably will if they ask you