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
