Question #92542

1 Answer
May 14, 2016

See explanation...

Explanation:

Note that vertex form of a vertical parabola can be written:

y = a(x-h)^2+ky=a(xh)2+k

where aa is a multiplier determining the 'steepness' of the parabola and (h, k)(h,k) is the vertex (turning point).

We are given:

f(x) = 1-xf(x)=1x

g(x) = 2x^2-9g(x)=2x29

h(x) = (g @ f)(x)h(x)=(gf)(x)

= g(f(x))=g(f(x))

= g(1-x)=g(1x)

= 2(1-x)^2-9=2(1x)29

= 2(1-2x+x^2)-9=2(12x+x2)9

= 2-4x+2x^2-9=24x+2x29

= color(blue)(2x^2-4x-7)=2x24x7

= 2(x^2-4x+4)-15=2(x24x+4)15

= 2(x-2)^2-15=2(x2)215

= color(blue)(2(x+(-2))^2+(-15))=2(x+(2))2+(15)

The minimum value of h(x)h(x) occurs when (x-2)^2 = 0(x2)2=0. That is when x=2x=2. hence the turning point is at color(blue)(""(2, -15))(2,15)

If k(x)k(x) is a translation of h(x)h(x) then it must have the same multiplier 22. Given that its turning point is at (-1, -5)(1,5), its equation can be written:

color(blue)(k(x) = 2(x+1)^2-5)k(x)=2(x+1)25