difficult function question of p1

[YuriGagarin]

can anyone solve tell me how to solve the last part of this question

 

[mrpaudel]

iii) f(x)=h(x-2)
or,(x-2)^2+3=h(x-2)
Let a=(x-2) then, a^2+3=h(a).....so h(x)=x^2=3

 

[YuriGagarin]

i didn't get the last part

 

[mrpaudel]

we have got h(x-2)=(x-2)^2+3.....
now accordin to this equation, if we substitute r=x-2 then, h(r)=r^2+3 ....so we are asked to get the value of h(x) so...replacin r by x in h(r) equation,
h(x)=X^2+3...got it...even if u dont get it..feel free to ask!!

 

[aliya_zad]

f(x)=hg(x)

u knw that gx is 2x-3 bt u dnt knw hx.
Consider hx to be ax2+bx+c.
Substitute gx in the hx equation.
x^2-4x+7= a(x-2)^2+b(x-2)+c
expand
x2-4x+7=ax2-4ax+4a+bx-2b+c

now equate the co-efficients.
ax2=x2
Therefore a=1

-4x=-4ax+bx
-4=-4+b
Therefore b=0

7= 4a-2b+c
7=4-0+c
c=3

equation =1x2+0x+3=(X2+3)

 

[mrpaudel]

At aliya Zad.....look at the question...this is a 1 mark question..so this is not gud spendin longer time jus for a mark..anyway..appreciate ur help!!

 

[YuriGagarin]

thnks guys

 

[mrpaudel]

Welcome....!! And if u r appearin this time....best of luck!!

 

[aliya_zad]

I knw mine is a long way but didnt know how to solve it this easily.
Thanks mrpaudel. It sure helps for 1 mark.