suppose that (3,6) is on the graph of y=j(x). Find the point on the graph of the given function with the given transformations y=3j(1/3x-2)-4

Answer 1

Pretty confusing. It may be easier, at least it is for me, to invent a function j(x) that goes through (3.6). One that is easy to follow is if you make a parabola with the vertex at (3,6) so in vertex form, that would be:

j(x) = (x - 3)² + 6 Then we can make the changes and see where the new vertex is.

Multiplying this out, we get: x² - 6x + 9 + 6 or x² - 6x + 15. Now make the transformations:

j(1/3x - 2) = (1/3x - 2)² - 6(1/3x - 2) + 15 = 1/9x² - 4/3x + 4 - 2x + 12 + 15 = 1/9x² - 4/3x - 6/3x + 31 or

j(1/3x - 2) = 1/9x² - 10/3x + 31

Now multiply by 3 to get 3[j(1/3x - 2)]

3[j(1/3x - 2)] = 1/3x² - 10x + 93

Now subtract 4 to get:

3[j(1/3x - 2)] - 4 = 1/3x² - 10x + 89

Calculate the vertex: x = -b/2a = 10/(2/3) = 10•(3/2) = 15. To find y, plug in x = 15 to get y = 14

So the point that was (3, 6) is now (15, 14)

Answer 2

Translate 2 units to the right to obtain y = j(x - 2) (3,6) → (5,6)

Stretch horizontally by a factor of 3 to obtain y = j ((1/3)x - 2) (5, 6) → (15, 6)

Stretch vertically by a factor of 3 to obtain y = 3j( (1/3)x - 2) (15, 6) → (15, 18)

Translate 4 units downward to obtain y = 3j( (1/3)x - 2) - 4 (15, 18) → (15, 14)

So, (3, 6) gets mapped to (15, 14).

Answer 3

We are changing the point (3,6) by using the equation y=3j(1/3x-2)-4. We start at the inside of the parenthesis at 1/3x-2 and substitute x for 3. 1/3(3)-2=1-2 or -1. Multiply -1 by 3j to have -3j. Then subtrade the -4. Thus the x component is x and the y component becomes -3j-4. the new point is ( 3, -3j-4).