Question

Set up an equation of a tangent to the graph of the following function.
$y=4x-x^2$ at the points of its intersections with the Ox axis.

Answer 1

The point where $y=4x-x^2$ meets x-axis has y-coordinate zero

$0=4x-x^2$

$x=0,4$

If $x=0,y=0$

If $x=4,y=0$

$\frac{dy}{dx}=4-2x$

${\left(\frac{dy}{dx}\right)}_{x=0}=4=m_1$

${\left(\frac{dy}{dx}\right)}_{x=4}=-4=m_2$

The equation of tangent passing through (0,0) and having slope as 4 is

$y=4x$

The equation of tangent passing through (4,0) and having slope as -4 is

$y=-4(x-4)$

$4x+y-16=0$