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