Question

what is the base/original function of y=-2sqrt3x-12 -5?

Answer 1

\[y=-2\sqrt{3x-12}-5\]

Answer 2

\(y=\sqrt x\), ...or maybe... \(y=-\sqrt x\)

Answer 3

thank you! I was also asked to describe the series of transformations from the base function to \[y=-2\sqrt{3x-12}-5\] do you have any ideas with the answer?

Answer 4

\[y=-2\sqrt{3\left( x-4 \right)}-5\]

if you start with \(y=\sqrt x\)

> reflect across the x-axis, to get: \(y=\color{green}-\sqrt x\)
> stretch vertically by 2: \(y=-\color{green}2\sqrt x\)
> shrink horizontally by 3: \(y=-2\sqrt {\color{green}3x}\)
> translate right by 4: \(y=-2\sqrt {3(x\color{green}{-4})}=-2\sqrt {3x-12}\)
> translate down by 5: \(y==-2\sqrt {3x-12}\color{green}{-5}\)

Answer 5

wow this is so helpful! Can you guide me through writing this equation in the form of \[y=af(k(x-c))+d\]?

Answer 6

@PaxPolaris

Answer 7

start with \(y=f(x)\)
> if and only if \(a\) is negative, reflect across the x-axis, to get: \(y=-f(x)\)
> stretch vertically by |a| : \(y=af(x)\)
> if and only if \(k\) is negative, reflect across the y-axis: \(y=af(-x)\)
> shrink horizontally by k : \(y=af(kx)\)
> translate (move) horizontally by +c: \(y=af\left(k\left(x-c\right)\right)\)
> translate (move) vertically by +d: \(y=af\left(k\left(x-c\right)\right)+d\)

Answer 8

always do translations after you do stretch/shrink/reflect

Answer 9

for the y=af(k(x-c))+d part, I'm still not quiet sure with the steps, can you show me the numbers and variables from the given function?

Answer 10

@PaxPolaris

Answer 11

you mean in:\[y=-2\sqrt{3\left( x-4 \right)}-5\]
\[f(x)=\sqrt x\]\[a=-2\]\[k=3\]\[c=4\]\[d=-5\]

Answer 12

yep, how do you write \[y=-2\sqrt{3x-1}\] in the function notation y=af(k(x-c))+d?

Answer 13

@PaxPolaris

Answer 14

\[=-2\sqrt{3\left( x-\frac13 \right)}\]

\[f(x)=\sqrt x\]\[a=-2\]\[k=3\]\[c=1/3\]\[d=0\]

Answer 15

what about y = -2f[3(x - 4)] - 5 ?

Answer 16

@PaxPolaris

Answer 17

that is what we did for the original question

Answer 18

\[y=-2\sqrt{3\left( x-4 \right)}-5\]