math question...........?

the equation of the tangent line to f(x) at (a,f(a)) is y − f(a) = f′(a)(x − a), due to the fact the tangent line and the function have the same slope at the tangent point.

find f′(x) via first principles

lim h→0 f′(x) = [f(x + h) − f(x)]/h

f′(x) = [(-(x + h)^2 + 3(x +h) − 1) − (-x^2 + 3x − 1)]/h
∴ f′(x) = [(-(x^2 + 2hx + h^2) + 3(x +h) − 1) − (-x^2 + 3x − 1)]/h
∴ f′(x) = [-x^2 − 2hx − h^2 + 3x + 3h − 1 + x^2 − 3x + 1]/h
∴ f′(x) = [ − 2hx − h^2 + 3h]/h
∴ f′(x) = h[3 − 2x − h]/h
∴ f′(x) = 3 − 2x − h ; h ≠ 0
lim h→0 f′(x) = 3 − 2x

a = x = 1 then f(a) = 1 and f′(a) = 1
tangent line y − 1 = 1(x − 1)
∴ y = x <== tangnet line in slope-intercept form

f(1) = -(1)^2 + 3(1) - 1 = 3 is the y-coordinate
Ddifferentiaate
f'(x) = -2x + 3
When x = 1
f'(1) = -2(1) + 3 = -2 + 3 = 1 Hence the slope is '1'.
Hence the tangent line is
y - 3 = (1)(x - 1)
y - 3 = x - 1
y = x + 2
The Equation of the Tangent line.

f(x)=-x^2+3x-1=>
f '(1)=[-(1+h)^2+3(1+h)-1-(-1+3-1]/h as h->0
=>
f '(1)=[-1-2h-h^2+3+3h-2]/h as h->0
=>
f '(1)=[h-h^2]/h as h->0
=>
f '(1)=1

x=1=> f(1)=1
=> the equation of the tangent line is
y-1=1(x-1)
=>
y-x=0

answer is 5