What is the equation of the line with slope $3$ which is tangent to the curve $f(x)=7x-x^2$ ?

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Answer 1 · 2016-09-01T12:08:48

$y=3x+4$

If $f(x)=7x-x^2$
then the slope (for any general $x$ value) is $f'(x)=7-2x$

when the slope is $m=f'(x)=3$
then $7-2x=3$
$color(white)("XX")-2x=-4$
$color(white)("XX")x=2$

If $x=2$ then
$color(white)("XXX")f(color(red)(2))=7*(color(red)(2))-color(red)(2)^2=14-4=color(blue)(10)$

and the point on the curve were $f'(x)=3$ occurs at $(color(red)(2),color(blue)(10))$

Therefore, for the tangent, we have a slope of $color(green)m=3$ and a point $(color(red)(2),color(blue)(10))$

Using the slope-point form of the equation:
$color(white)("XXX")y-color(blue)(10)=color(green)(3)(x-color(red)(2))$

or
$color(white)("XXX")y=3x+4$

What is the equation of the line with slope 33 which is tangent to the curve f(x)=7x-x^2f(x)=7x-x^2?