Differentiation find equation of the normal

[ka923]

I have completed 3a and my answer was f'(x)=5/2x^-1/2 + 12x^-2
my progress so far on 3B:
I have subbed in x=4 in my original equation of f(x)
which gives me 5root4 - 12/4 =7
so y =7
then i subbed in 4 for f'(x) which gave me m=2
i then subbed in all the information i have into y -y1 = m(x-x1)
which is y-7 = 2(x-4)
= 2y-14=x-4
this is what im up to right now but i believe that im wrong as the mark scheme states 2y+x=18 should be my answer
please help thanks

 

[Harry_the_cat]

View attachment 10583
I have completed 3a and my answer was f'(x)=5/2x^-1/2 + 12x^-2
my progress so far on 3B:
I have subbed in x=4 in my original equation of f(x)
which gives me 5root4 - 12/4 =7
so y =7
then i subbed in 4 for f'(x) which gave me m=2 … all good up to here, but this is the gradient of the tangent. You need the gradient of the normal.
i then subbed in all the information i have into y -y1 = m(x-x1)
which is y-7 = 2(x-4)
= 2y-14=x-4
this is what im up to right now but i believe that im wrong as the mark scheme states 2y+x=18 should be my answer
please help thanks

see comments in red

 

[ka923]

ahh i get it now thank you very much