# Thread: Simple graph question (conflicting solution)

1. ## Simple graph question (conflicting solution)

Now, I have answered this question and it seems very simple. But the answer they give is different to mine, and i graphed to check, and it seems I am correct, but I want you guys to tell me what you think. Part c) is what I'm concerned with.

a) The graph of g is obtained from the graph of the function f with rule $\displaystyle f(x) = x^2$ by a translation by +3 units parallel to the x-axis. Write down the rule for g.

Now, i wrote $\displaystyle (x-3)^2$ which is correct.

b) The graph of h is obtained from the graph of g by a translation by −1 unit parallel to the y-axis. Write down the rule for h.
I wrote $\displaystyle (x-3)^2-1$

c) The graph of k is obtained from the graph of h by a dilation by a scale factor of 0.5 from the y-axis. Write down the rule for k.

I wrote $\displaystyle 2(x-3)^2-1$, but they say:
$\displaystyle (2x-3)^2-1$. Now, how can that be correct since their version changes the turning point completely? It's basically a totally different graph, isn't it?

So tell me what you think.

2. Originally Posted by scorpion007
Now, I have answered this question and it seems very simple. But the answer they give is different to mine, and i graphed to check, and it seems I am correct, but I want you guys to tell me what you think. Part c) is what I'm concerned with.

a) The graph of g is obtained from the graph of the function f with rule $\displaystyle f(x) = x^2$ by a translation by +3 units parallel to the x-axis. Write down the rule for g.

Now, i wrote $\displaystyle (x-3)^2$ which is correct.

b) The graph of h is obtained from the graph of g by a translation by −1 unit parallel to the y-axis. Write down the rule for h.
I wrote $\displaystyle (x-3)^2-1$

c) The graph of k is obtained from the graph of h by a dilation by a scale factor of 0.5 from the y-axis. Write down the rule for k.

I wrote $\displaystyle 2(x-3)^2-1$, but they say:
$\displaystyle (2x-3)^2-1$. Now, how can that be correct since their version changes the turning point completely? It's basically a totally different graph, isn't it?

So tell me what you think.
I think the given answer is correct. Let $\displaystyle f_1(x)$ be the function
you want to dilate. You want a new function $\displaystyle f_2(x)$ such that:

$\displaystyle f_2(x)=f_1(2x)$

Now let $\displaystyle f_1(x)=(x-3)^2-1$, then:

$\displaystyle f_2(x)=f_1(2x)=(2x-3)^2-1$

RonL