Question

Set up an equation of a tangent to the graph of the following function.
$y=x^2-2x$ at the points of its intersection with the abscissa axis.

Answer 1

$y=x^2-2x$

To find out the point where the curve intersects abscissa is $y=0$

$x^2-2x=0$

$x(x-2)=0$

$x=0,2$

$\frac{dy}{dx}=2x-2$

${\left(\frac{dy}{dx}\right)}_{x=0}=-2=m_1$

${\left(\frac{dy}{dx}\right)}_{x=2}=2=m_2$

$(x_1,y_1)=(0,0),(x_2,y_2)=(2,0)$

$(x-x_1)m_1=y-y_1$

$2x+y=0$ is the tangent at $(0,0)$

$(x-x_2)m_2=y-y_2$

$(x-2)2=y-0$

$2x-y-4=0$ is the tangent at $(2,0)$