I am a year 11 student getting E's in EXT1 maths but I will give this question a go.

f'(x)=4x-3

f(2)=7

find(4)

we know that f(x)=2x^2 -3x

because to find f' we times the power to the coefficient and reduce the power by 1.

so lets test if f(2)=4

4=2(2)^2-3(2)

we get :

4=2 which is wrong

now we know that in quadratic equation there can be a constant which determines if vertex is down or up

so lets make that constant 'c'

so we get the equation

7=2+c

c=5

now we know that the correct equation is f(x)= 2x^2-3x+5

find f(4)

let y be f(4)

y=2(4)^2-3(4)+5

=25

Therefore f(4)=25