 Hello friends, and how are you all doing today? The question says find the equation of all lines having slope 0 Which are tangent to the curve y is equal to 1 upon x square minus 2x plus 3 Let us rewrite the equation of the curve once again Now here Let's first differentiate y with respect to x and on doing so we get dy by dx is equal to minus 1 into 2x minus 2 upon x square minus 2x plus 3 the whole square that is on Using the quotient rule on Simplifying it further we have 2 minus 2x upon x square minus 2x plus 3 the whole square as Our dy by dx also. We are given that the slope That is dy by dx is Equal to 0 so we have 0 is equal to 2 minus 2x upon x square minus 2x plus 3 the whole square or 2 minus 2x is equal to 0 or x is equal to 1 now when the value of x is equal to 1 the value of y is 1 upon x square That is 1 minus 2x that will be 2 plus 3 That gives us the value of y as 1 by 2 so the point of of contact is 1 comma 1 by 2 Now we need to find out the equation of line through This point that is 1 1 by 2 and having a slope as 0 we have y minus y 1 Equal to n that is slope bracket x minus x 1 Or we have 2y minus 1 upon 2 equal to 0 Or we have y is equal to 1 by 2 So the answer to the question given to us is The equation of the line is y is equal to 1 by 2 Right, so this completes the session. Hope you understood it. Have a nice day