[Super Mario 64]

the graph y=e^x is transformed to y=e^kx

answer is stretch of 1/k in x direction.

But why is it not a stretch of e^k? since e^x multiplied by e^k is e^kx

[raheem94]

(Original post by Super Mario 64)
the graph y=e^x is transformed to y=e^kx

answer is stretch of 1/k in x direction.

But why is it not a stretch of e^k? since e^x multiplied by e^k is e^kx

[kontemptXD]

(Original post by raheem94)

Can you explain why it is 1/K please ?

[The Polymath]

(Original post by Super Mario 64)
the graph y=e^x is transformed to y=e^kx

answer is stretch of 1/k in x direction.

But why is it not a stretch of e^k? since e^x multiplied by e^k is e^kx

Oh I remember coming up with a way to explain transformations I'll simplify it with y = kx as it's the same in principle.

Basically, with a normal y=x graph, y takes on it's "correct" value straight away. With y = 2x, the y value reaches the normal y=x value twice as quickly, meaning the graph is compressed horizontally.

e.g. with y = x, y = 4 when x = 4, but with y = 2x, y reaches 4 when x = 2 instead, which is twice as 'quickly' if you try to consider the x-axis as a sort of timescale.

Equally with y = x + 1, y reaches the value of 4 one "x-unit" early, meaning the graph shifts to the left.

Hopefully that makes sense, it certainly helped me to get my head round it.

[raheem94]

(Original post by kontemptXD)
Can you explain why it is 1/K please ?

So this represents a horizontal stretch of scale factor

[Super Mario 64]

(Original post by Junaid96)
Oh I remember coming up with a way to explain transformations I'll simplify it with y = kx as it's the same in principle.

Basically, with a normal y=x graph, y takes on it's "correct" value straight away. With y = 2x, the y value reaches the normal y=x value twice as quickly, meaning the graph is compressed horizontally.

e.g. with y = x, y = 4 when x = 4, but with y = 2x, y reaches 4 when x = 2 instead, which is twice as 'quickly' if you try to consider the x-axis as a sort of timescale.

Equally with y = x + 1, y reaches the value of 4 one "x-unit" early, meaning the graph shifts to the left.

Hopefully that makes sense, it certainly helped me to get my head round it.

Thanks for the explanation! But wouldn't y=x+1 mean it shifts 1 unit in the positive Y direction?

[The Polymath]

(Original post by Super Mario 64)
Thanks for the explanation! But wouldn't y=x+1 mean it shifts 1 unit in the positive Y direction?

Same thing in this case, as f(x+1) = f(x) +1 (for this particular line) If I tok y = (x+1)^2 then it would be a shift 1 to the left, definitely not up.