Question

Will medal: How would I vertically stretch 4x^2+3x by a factor of 2?

Answer 1

stretching a curve vertically sounds like you will make all the "stretched" values of y bigger than they would be if they weren't stretched, right?

So if you want a y-value of 10 to be stretched vertically by a factor of 2, you need the new y-value to be 20, basically twice as big.

In this problem, you have y = 4x^2+3x

And for any x, you get a y.... but you want a stretched y that is twice as big. What happens to the y values if you multiply the whole x expression by 2?

Answer 2

I see how I can stretch something like x^2 by a factor of 5. I would just do 5x^2.. right?

I'm not sure what to do when it's like a binomial, such as above (4X^2+3x)

Answer 3

8x^2+6x, is this correct?

Answer 4

ok, good.. you're practically there...

if y = 4x^2 + 3x

let z be the stretched version of y.... z = 2y

so z = 2y = 2(4x^2 +3x) = 8x^2 + 6x

Answer 5

just like you multiplied by 5 in your example, you multiply by 2 here, but you have to distribute the 2 across all the terms in the expression.

Answer 6

Alright, thank you so much!

Answer 7

Here's a point to convince you...

y = 4x^2+3x (x = 1, y = 7)
z = 8x^2 + 6x (x = 1, z = 14)

Answer 8

glad to help... you probably didn't need convincing, but seeing it in action can help it make sense... Good luck!

Answer 9

So 4x^2+3x stretched vertically by a factor of 2 would be 8x^2+6x, because I distribute the two across the board.

Answer 10

right!

Answer 11

Have you done horizontal stretch yet?

Answer 12

Yes, that is where I have to be careful, correct?
So if I want to stretch horizontally by a factor of 4, I want everything to be 1/4, right?

Answer 13

careful is right... I was starting to wish I had thought about it before asking ;)

if you want to horizontally stretch, then the value of the function y for the original x is just y. The value of the function by a factor of 4 will still be y, but it will occur when you input a value for x that is 4 times as large.

Answer 14

y = 4x^2+3x point: (x = 1, y = 7)

z = 4(x/4)^2 + 3(x/4) point: (x = 4, y = 7)

of course, you might need to simplify the expression, but yes, every "new x" needs to be "x/4"

Answer 15

Oh, I see! Thank you! Just replace x with x/4, that's actually really easy. (And I thought the horizontal one was hard!

Answer 16

well... me too :) That's why I like finding a point or two on the original curve and on the stretched curve.. makes sure I'm not making a dumb error!

Answer 17

the goal of all this is to know, for any function, how to stretch, compress, and shift up/down/left/right.