Function Transformations

[Gragadoodle]

Why is f(x) = 1/2 (x^3) a vertical shrink?

Isn't is supposed to be a horizontal stretch, because f(x) = f(cx), 0<c<1?

 

[lookagain]

Why is f(x) = (1/2) (x^3) a vertical shrink?

Isn't is supposed to be a horizontal stretch, because f(x) = f(cx), 0<c<1?

It has been compressed (or shrunk) by a factor of 1/2. The coefficient of 1/2 has made the base function less steep everywhere. Its graph has become more flattened.

 

[Gragadoodle]

Yes, but why isn't it a horizontal stretch?

If there is a coefficient of greater than 0 but less than 1 multiplied by x it's always a horizontal stretch.

 

[Gragadoodle]

Bump

 

[daon2]

Bump

What I wrote was incorrect, an intuition fail. Horizontal stretching occurs when the argument is changed by a multiple of itself. \(\displaystyle f(\frac{1}{2}x)\) will change \(\displaystyle f(x)\) by impeding its rate of change. For example at \(\displaystyle x=2\) \(\displaystyle f(x)=f(2)\) but \(\displaystyle f(\frac{1}{2}x)=f(1)\), where \(\displaystyle f(x)\) used to be half the graph ago.

 

[Gragadoodle]

I'm looking at this in a mathematical point of view.

Technically, it's a vertical shrink not a horizontal stretch.

 

[Gragadoodle]

What I wrote was incorrect, an intuition fail. Horizontal stretching occurs when the argument is changed by a multiple of itself. \(\displaystyle f(\frac{1}{2}x)\) will change \(\displaystyle f(x)\) by impeding its rate of change. For example at \(\displaystyle x=2\) \(\displaystyle f(x)=f(2)\) but \(\displaystyle f(\frac{1}{2}x)=f(1)\), where \(\displaystyle f(x)\) used to be half the graph ago.

Horizontal stretching occurs when y= f(cx), 0<c<1

 

[daon2]

Horizontal stretching occurs when y= f(cx), 0<c<1

So this is both a horizontal stretch and a vertical compression of \(\displaystyle f(x)=x^3\).

\(\displaystyle \frac{1}{2}f(x) = \frac{1}{2}x^3 = \left(\dfrac{1}{\sqrt[3]{2}} x\right)^3 =f\left(\dfrac{1}{\sqrt[3]{2}} x\right)\)

 

[lookagain]

Why is f(x) = (1/2) (x^3) a vertical shrink?

Isn't is supposed to be a horizontal stretch, because f(x) = f(cx), 0<c<1?It has been vertically compressed (or shrunk) by a factor of 1/2. \(\displaystyle \ \ \) <--- edit

The coefficient of 1/2 has made the base function less steep everywhere. Its graph has become more flattened.

I left out the required word of "vertically."