Question

Help! Algebra! Will fan and medal

Answer 1

Given that f(x) = x2 + 3x + 6 and g(x) = the quantity of \[\frac{ x-3 }{ 2 }\] , solve for f(g(x)) when x = 1.

Answer 2

possible answers:
-1
0
3

Answer 3

and 4

Answer 4

\[ f(\color{red}{x})=\color{red}{x}^2+3\color{red}{x}+6\\
f(\color{red}{g(x)})=(\color{red}{g(x)})^2+3(\color{red}{g(x)})+6\]

Substitute g(x) for your given expression into the equation.

Then to get f(g(x)) when x =1, use the above result and substitute x=1 wherever you see x

Answer 5

Okay, I wrote that all down... So next, solve for f(x) ?

Answer 6

plug in x=1 into your f(g(x)) expression

Answer 7

so the f(g(x)) equation equals 10

Answer 8

\[f(\color{red}{g(x)})=(\color{red}{g(x)})^2+3(\color{red}{g(x)})+6\]
since \(g(x)=\dfrac{x-3}{2}\), then:
\[f(\color{red}{g(x)})=\left(\color{red}{\dfrac{x-3}{2}}\right)^2+3\left(\color{red}{\dfrac{x-3}{2}}\right)+6 \]

Is that what you got? You just need to plug in x=1 now. It shouldn't give 10

Answer 9

I got 4.75

Answer 10

that's closer to the answer... but not quite.

Answer 11

Oh.. okay just give me a second to redo my work...

Answer 12

OHHH its 4