# Describe a sequence of transformations that transform the graph of f(x) into the graph of g(x)? f(x)=sqrtx and g(x)=-3(sqrt(x+1))-4

Dec 10, 2015

$\left(x , f \left(x\right)\right) \rightarrow \left(x - 1 , \sqrt{x}\right) \rightarrow \left(x - 1 , - 3 \sqrt{x}\right) \rightarrow \left(x - 1 , g \left(x - 1\right)\right)$

#### Explanation:

1) Translation. x' := x - 1 ; y' = y.
Just send $\left(x , \sqrt{x}\right) \rightarrow \left(x - 1 , \sqrt{x}\right)$ therefore $\left(0 , 0\right) \rightarrow \left(- 1 , 0\right)$
And $\left(1 , 1\right) \rightarrow \left(0 , 1\right)$

2) Stretch. x'' = x' ; y'' := -3y'
Just send $\left(x ' , y '\right) \rightarrow \left(x ' , - 3 y '\right)$ therefore $\left(0 , 1\right) \rightarrow \left(0 , - 3\right)$
This is a dilatation (multiply by 3)
followed by a refraction (multiply by -1). The mirror is x-axis.

3) Translation. x''' := x'' ; y''' := y'' - 4.
Just send $\left(x ' ' , y ' '\right) \rightarrow \left(x ' ' , y ' ' - 4\right)$
Therefore $\left(- 1 , 0\right) \rightarrow \left(- 1 , - 4\right)$
And $\left(0 , - 3\right) \rightarrow \left(0 , - 7\right)$

Dec 11, 2015

$f \left(x\right) = \sqrt{x}$
$\left(16 , 4\right)$
graph{sqrtx [-1.705, 18.295, -3.44, 6.56]}

$a \left(x\right) = \sqrt{x \textcolor{red}{+ 1}}$
Function shifts one to the left.
$\left(15 , 4\right)$
graph{sqrt(x+1) [-2.16, 17.84, -2.08, 7.92]}

$b \left(x\right) = \textcolor{red}{3} \left(\sqrt{x + 1}\right)$
Function is stretched vertically by a factor of $\text{3}$.
$\left(15 , 12\right)$
graph{3sqrt(x+1) [-2.08, 33.48, -2.43, 15.35]}

$c \left(x\right) = \textcolor{red}{-} 3 \left(\sqrt{x + 1}\right)$
Function is reflected across the $x$-axis.
$\left(15 , - 12\right)$
graph{-3sqrt(x+1) [-3.42, 32.14, -15.24, 2.54]}

$g \left(x\right) = - 3 \left(\sqrt{x + 1}\right) \textcolor{red}{- 4}$
Function is moved $\text{4}$ units down.
$\left(15 , - 16\right)$
graph{-3sqrt(x+1)-4 [-4.75, 35.25, -16.85, 3.15]}