A quadratic function y=f(x)  if it touches the line y=x at the point x=1 and   passes through the point (-1, 0)

y=f(x)=ax2+bx+cf′(x)=2ax+b=1 at x=1⇒2a+b=1----1x=1_,_y=1⇒f(1)=1⇒a+b+c=1----2 Solving (1)&2 curve passes through (−1,0)a=14b=12c=14Fx==x24+x2+14=x2+2x+14=14(x+1)2