Question

@sidsiddhartha

Answer 1

given
\[f(x)=4.5x^2-3x+2\]
The problem gives you the 'x' at 2
now,Find the 'y' by find the value of f(2):
\[f(2)=4.5*2^2-3*2+2=14\]
so till now we have x=2,y=9 right?

Answer 2

sorry x=2 and y=14

Answer 3

now find the derivative or the slope
\[f'(x)=9x-3\\now~f'(2)=9*2-3=15=slope(m)\]

Answer 4

now simply use
\[y-y_1=m(x-x_1)\\y-14=15(x-2)\\15x-y=16\]
thats the equation of the tangent at x=2
@mondona