Question

The tangent at A (2, 4) on $y=x^3-2x^2+4$ cuts the x axis at T then AT =

• A. $4\sqrt{17}$
• B. $\frac{\sqrt{17}}{4}$
• C. $\sqrt{17}$
• D. $17$

Answer 1

The correct option is C $\sqrt{17}$

$\frac{dy}{dx}=m=3x^2-4x=12-8=4$
​AT =$\left|\frac{y_1\sqrt{1+m^2}}{m}\right|=\frac{4.\sqrt{17}}{4}=\sqrt{17}$