Find Vertex of h(x) After Translations & Reflection

[Senjai]

Homework Statement

Let g(x) be the quadratic function:
g(x) = (x - 1)^2 + 2

A) Suppose h(x) is the result of g(x) undergoing a translation of 5 units to the right and 4 units up, and then a reflection over the y axis. find the coordinate of the vertex of h(x).

B) A Transformation involving vertical and horizontal scale factors only

h(x) \rightarrow ah(bx)

will bring the vertex of h(x) back to the vertex of g(x). Find the values of a and b (the rest of the function will *not* be the same as g(x)).

Homework Equations

y = af[b(x-h)] + k [/tex]

The Attempt at a Solution

A) First i declared my translations:

x -> x-5
y -> y -4

so far my function looks like this.

y - 4 = (x-1-5)^2 + 2
y = (x-6)^2 + 6

now we have a reflection over the y axis.

x -> -x, now my function looks like this. h(x) = (-x-6)^2 +6

So the vertex must be where h(x) = 6, so i sub y in for 6.
and solve for x

6 = (-x-6)^2+6
and i get x = -6, so the vertex would be (-6, 6), graphing calculator confirmed this.

B) this is where i encountered issues.
a and b equal the same as h(x) because its just a reflection, in
y = ah(b(x-h))+k i declared b = -1, and a = 1, is this the right way to show my answer? but just to show how i would get the vertex back to g(x)'s vertex, i also stated the values for h and k, h = -5, k = -4. and stated that g(x) = h(-x+5) - 4, is this correct?

 

[olgranpappy]

Homework Statement

Let g(x) be the quadratic function:
g(x) = (x - 1)^2 + 2

A) Suppose h(x) is the result of g(x) undergoing a translation of 5 units to the right and 4 units up, and then a reflection over the y axis. find the coordinate of the vertex of h(x).

B) A Transformation involving vertical and horizontal scale factors only

h(x) \rightarrow ah(bx)

will bring the vertex of h(x) back to the vertex of g(x). Find the values of a and b (the rest of the function will *not* be the same as g(x)).

Homework Equations

y = af[b(x-h)] + k [/tex]

The Attempt at a Solution

A) First i declared my translations:

x -> x-5
y -> y -4

so far my function looks like this.

y - 4 = (x-1-5)^2 + 2
y = (x-6)^2 + 6

now we have a reflection over the y axis.

x -> -x, now my function looks like this. h(x) = (-x-6)^2 +6

So the vertex must be where h(x) = 6, so i sub y in for 6.
and solve for x

6 = (-x-6)^2+6
and i get x = -6, so the vertex would be (-6, 6), graphing calculator confirmed this.

B) this is where i encountered issues.
a and b equal the same as h(x) because its just a reflection,

The above statement doesn't make sense (to me, at least)...

Anyways, after cleaning up a little bit, you have:
<br /> h(x)=(x+6)^2+6\;.<br />

What is
<br /> h(bx)<br />
equal to?

 

[Senjai]

in the second part of the question (kind of jumps back to the top, sorry) it asks for the values of a and b, in order to translate h(x)'s vertex back to g(x)'s
h(x) \rightarrow ah(bx)
where a is the vertical scale factor, and b is the horizontal scale factor.

 

[boaz]

i haven't tried to solve it deeply but as much as i can see from here your way is fine. due to the fact that you have had 2 equations (y and its derivative) and 4 variables, you have had no choice but to set values for h and k in order to get the values of a and b.

 

[olgranpappy]

in the second part of the question (kind of jumps back to the top, sorry) it asks for the values of a and b, in order to translate h(x)'s vertex back to g(x)'s
h(x) \rightarrow ah(bx)
where a is the vertical scale factor, and b is the horizontal scale factor.

Right... but, I'd like to know whether or not you can tell me:

Given
<br /> h(x)=(x+6)^2+6\;,<br />
then
what is
<br /> h(bx)<br />
equal to?