Graphs and Transformations

Why is the transformation that maps the graph of y = 4x^2+3 onto y= x^2 + 3, equal to stretch parallel to the x axis with scale factor 2?

Here's my workings:
Change only being applied to x so its along the x axis
It is a division of 4 so sf.0.25

Why is this wrong?

Reply 1

β0b

21

Original post by Hollymae764

Why is the transformation that maps the graph of y = 4x^2+3 onto y= x^2 + 3, equal to stretch parallel to the x axis with scale factor 2?

Here's my workings:
Change only being applied to x so its along the x axis
It is a division of 4 so sf.0.25

Why is this wrong?

picture pleeease.

Reply 2

β0b

21

Original post by β0b

picture pleeease.

wait no, gimme a sec.

Reply 3

β0b

21

Original post by β0b

wait no, gimme a sec.

I used desmos. it does get larger, which makes sense, as it is inside the function meaning it does the opposite. because you are dividing by 4, it must be multiplied by 4. the one thing which gets me is the scale factor of 2.

Reply 4

β0b

21

the blue is the second the red is the first.

desmos-graph.png73.8KB

Reply 5

Hollymae764

OP

12

Im still confused is there an algebraic rule you have to follow?

Reply 6

nzy

17

When you stretch a graph along the x-axis, you would usually have to consider a graph of y=f(ax), or in this case, y=(ax)^2 + 3. The mutiplier a strictly applies to the value of x, not x^2. Therefore, y = 4x^2 + 3 should first be written as y = (2x)^2 + 3 before you can compare it to y = x^2 + 3. With that you should be able to deduce the correct answer.

Reply 7

β0b

21

Original post by nzy

When you stretch a graph along the x-axis, you would usually have to consider a graph of y=f(ax), or in this case, y=(ax)^2 + 3. The mutiplier a strictly applies to the value of x, not x^2. Therefore, y = 4x^2 + 3 should first be written as y = (2x)^2 + 3 before you can compare it to y = x^2 + 3. With that you should be able to deduce the correct answer.