f x x2 then f x the tangent to the graph of f through th
f (x) = x^2 then f \' (x) = ? the tangent to the graph of f through the point (1,1) has the slope .......? and the y-intercept ........? it intercepts with the x-axis at x = ?
Solution
f(x)= x^2
 
 f\'(x)= 2x
 
 at(1,1)
 slope =2 x= 2
 
 so, equation of tangent is
 y-1 = 2(x-1)
 => y= 2x-1
 so, y-intercept is -1
 
 put y=0 => x=0.5
 => x-intercept = 0.5
 meets x-axis at (0.5,0)

