A standard function form (universal formula) for translations and dilations is y=kf(a(x+b))+c where the image point is transformed from (x,y) into (a(x+b),ky+c).
To find the coordinates of the image point when the function is transformed into y=6f(x+1)+8, identify that k=6, b=-1, c=8.
Now transform the image point using y=kf(a(x+b))+c where the image point is transformed from (x,y) into (a(x+b),ky+c), and (5,-2) is the original point, k=6, b=-1, c=8.
The new image point is (1(5-1),(6)(-2)+(8)) which simplifies to (4,-4).
(4,-4)
Question 2 of 3
2. Question
(3,-2) lies on y=f(x). Find the coordinates of the image point when the function is transformed into y=4f(-x)-9.
A standard function form (universal formula) for translations and dilations is y=kf(a(x+b))+c where the image point is transformed from (x,y) into (a(x+b),ky+c).
To find the coordinates of the image point when the function is transformed into y=4f(-x)-9, identify that k=4, a=-1, b=0, and c=-9.
Now transform the image point using y=kf(a(x+b))+c where the image point is transformed from (x,y) into (a(x+b),ky+c), and (3,-2) is the original point, k=4, a=-1, b=0, and c=-9.
The new image point is (-1(3+0),(4)(-2)+(-9)) which simplifies to (-3,-17).
(-3,-17)
Question 3 of 3
3. Question
(12,-1) lies on y=f(x). Find the coordinates of the image point when the function is transformed into y=-5f(3x-6)-4.
A standard function form (universal formula) for translations and dilations is y=kf(a(x+b))+c where the image point is transformed from (x,y) into (a(x+b),ky+c).
To find the coordinates of the image point when the function is transformed into y=-5f(3x-6)-4, identify that k=-5, a=3, b=-6, and c=-4.
Now transform the image point using y=kf(a(x+b))+c where the image point is transformed from (x,y) into (a(x+b),ky+c), and (12,-1) is the original point, k=-5, a=3, b=-6, and c=-4.
The new image point is (-4((3)(12)+-6),(-5)(-1)+(-4)) which simplifies to (6,1).