Find the equation of all lines of slope zero and that is tangent to th

1.	Find the equation of all lines of slope zero and that is tangent to the curve\(y=\frac{1}{x^2-2x+3}.\)
Answer» finding the slope of the tangent by differentiating the curve \(\frac{dy}{dx}=\frac{(2x-2)}{(x^2-2x+3)}\) Now according to question, the slope of all tangents is equal to 0, so \(-\frac{(2x-2)}{(x^2-2x+3)}=0\) Therefore the only possible solution is x = 1 since this point lies on the curve, we can find y by substituting x \(y=\frac{1}{1-2+3}\) \(y=\frac{1}{2}\) equation of tangent is given by y – y₁ = m(tangent)(x – x₁) \(y-\frac{1}{2}=0(x-1)\) \(y=\frac{1}{2}\)

Find the equation of all lines of slope zero and that is tangent to the curve\(y=\frac{1}{x^2-2x+3}.\)

Answer»

finding the slope of the tangent by differentiating the curve

\(\frac{dy}{dx}=\frac{(2x-2)}{(x^2-2x+3)}\)

Now according to question, the slope of all tangents is equal to 0, so

\(-\frac{(2x-2)}{(x^2-2x+3)}=0\)

Therefore the only possible solution is x = 1

since this point lies on the curve, we can find y by substituting x

\(y=\frac{1}{1-2+3}\)

\(y=\frac{1}{2}\)

equation of tangent is given by y – y₁ = m(tangent)(x – x₁)

\(y-\frac{1}{2}=0(x-1)\)

\(y=\frac{1}{2}\)