for the longest while i've been working on this question can any 1 help me out here Code: f(x) = 3^(x + 1) and g(x) = 3-x + 3 . Find the point of intersection of the graphs of f and g by solving f(x) = g(x).
f(x) = 3^(x + 1) and g(x) = 3-x + 3 . Find the point of intersection of the graphs of f and g by solving f(x) = g(x).
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$\displaystyle 3^{x+1} = 3-x+3$ Is that it? g(x) seems strange...
sorry this is what the question said Code: 3^{x+1} = 3^{-x+3}
3^{x+1} = 3^{-x+3}
$\displaystyle 3^{x+1} = 3^{3-x}$ Insert log on both sides: $\displaystyle \log 3^{x+1} = \log 3^{3-x}$ $\displaystyle (x+1)\log 3 = (3-x)\log 3$ $\displaystyle (x+1) = (3-x)$ Can you finish now?
so am i solving for x using quadratic then what
You don't need to use quadratic...
so how am i going to get the points
You solve for x... $\displaystyle x+1 = 3-x$
3-x -x -1=0 -2x +2 = 0 x = -1 is it this what do i do after
Since you are looking for the point of intersection, you are looking for a coordinate. You already have the x coordinate, what can you do it this and the two equations: $\displaystyle f(x) =3^{x+1}$ $\displaystyle g(x) = 3^{x-3}$
so i solved the equation above and got -1 so am i going to substitute it in fx and gx and solve
Yes Substituting in either f(x) or g(x) should be enough though.
i guess i have to substitute for both bcuz they said for f and g for f i got (-1,1) any suggestions
Well... actually you didn't solve for x properly... I thought that you were doing the calculations well, but a small mistake crept in: -2x +2 = 0 -2x = -2 x = 1 Now, find your y coordinates.
cud u jus verify this for me 0 = 3-1-2x 0 = 2-2x -2= -2x x=1 f(x) = 3^(x+1) f(x) = 3^ (1+1) f(x)= 9 g(x) = 3^(-1+3) g(x) = 3^(2) g(x) = 9
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