Quadratic Transformation Help

Question

F(x) = -(2(x-4))^2+2

describe the transformations that would turn f(x) into quad function g(x) = -2f(3(x-4))+2

more specifically, how to find the vertex of g(x)

Answer 1

right shift 4 units, dilation by factor of 1/ 18, shift up 2 units and flip about the x-axis

f(x) = -(2(x-4))^2 + 2

= - (4(x-4)^2) + 2

= - (4(x^2 - 8x + 16)) + 2

= -4x^2 + 32x - 64 + 2

= -4x^2 + 32x - 62

G(x) = -2 f ( 3x - 12) + 2

= -2 { -4(3x-12)^2 + 32(3x-12) - 62}+ 2

= -2 { -4(9x^2 -72x + 144 + 96x - 384 - 62} + 2

= 8 (9x^2 -72x + 144 + 96x - 384 - 62} + 2

= 8 (9x^2 + 24x -300)

= 24( 3x^2 + 8x - 100)

= 72( x^2 + (8/3)x - 100/3)

= 72 ( x^2 + (8/3)x + 64/36) - 100/3 + 64*72/36 <--- completes the square

= 72 ( x + 4/3)^2 + 28

So the vertex is (-4/3, 28)