Question 470687


First {{{y = f(x) }}}and {{{f(x) = x^3 - 4x}}}, {{{y = g(x) }}}and {{{g(x) = f(x + h) + k}}}  or 

{{{y = f(x) = x^3 -4x}}}
{{{y = g(x) = f(x + h) + k}}}

Looks to me like we have two equations for {{{y}}} which means we can solve them as a system. Notice that {{{g(x) }}}contains {{{f(x)}}}  but with a different input variable, namely {{{(x + h)}}}. 

To find {{{f(x + h)}}} we replace{{{ x}}} with {{{(x + h)}}} wherever we find an {{{x}}} in the {{{f}}} equation.

{{{f(x) = x^3 - 4x}}} will become

{{{f(x + h) = (x + h)^3 - 4(x + h)}}}

So now we replace {{{f(x + h)}}} in {{{g(x)}}} with our new result.

{{{g(x) = f(x + h) +k}}}

{{{g(x) = ( (x + h)^3 - 4(x + h) ) + k}}}

Also, we are given a pair of points that work with {{{g(x)}}} so lets plot them in and get a pair of equations. The first number in each point is {{{x }}}and the second is {{{y}}}.

Points in g(x) : ({{{2}}}, {{{1}}}) and ({{{4}}}, {{{-5}}})

So for the first :

{{{1 = (2 + h)^3 - 4(2 + h) + k}}}

{{{1 = (8 + 12h + 6h^2 + h^3) - (8 + 4h) + k}}}

{{{1 = 8 + 12h + 6h^2 + h^3 - 8 - 4h + k}}}

{{{1 = h^3 + 6h^2 + 8h + k}}}

{{{1 - (h^3 + 6h^2 + 8h) = k}}}

{{{- h^3 - 6h^2 - 8h + 1 = k}}}

and for the second :

{{{-5 = (4 + h)^3 - 4(4 + h) + k}}}

{{{-5 = (64 + 48h + 12h^2 + h^3) - (16 + 4h) + k}}}

{{{-5 = 64 + 48h + 12h^2 + h^3 - 16 - 4h + k}}}

{{{-5 = h^3 + 12h^2 + 44h + 48 + k}}}

{{{-5 - (h^3 + 12h^2 + 44h + 48) = k}}}

{{{- h^3 - 12h^2 - 44h - 53 = k}}}

Notice that both long equations are equal to {{{k}}}. That means they are also equal to each other.

{{{- h^3 - 6h^2 - 8h + 1 = - h^3 - 12h^2 - 44h - 53}}}

{{{- h^3 - 6h^2 - 8h + 1 - (- h^3 - 12h^2 - 44h - 53) = 0}}}

{{{- h^3 - 6h^2 - 8h + 1 + h^3 + 12h^2 + 44h + 53 = 0}}}

{{{6h^2 + 36h + 54 = 0}}}

{{{6(h^2 + 6h + 9) = 0}}}

{{{(h + 3)(h + 3) = 0 / 6}}}

{{{h + 3 = 0}}}

{{{h = -3}}}

So{{{ h}}} is {{{-3}}}, now we need {{{k}}}.

{{{g(x) = ( (x + h)^3 - 4(x + h) ) + k}}}

{{{y = ( (x + h)^3 - 4(x + h) ) + k}}}

{{{1 =( (2 - 3)^3 - 4(2 - 3)) + k}}}

{{{1 = -1^3 - 4(-1) + k}}}

{{{1 = -1 + 4 + k}}}

{{{1 = 3 + k}}}

{{{-2 = k}}}

So {{{k }}}is {{{-2}}}. We already know that {{{h}}} is {{{-3}}}. so :

{{{hk = -3*-2=6}}}