Question 1205498

the equation of the tangent line to the curve 

{{{y = -x^2- 3x + 4}}} at {{{x= 2}}}

 The gradient of the tangent is equal to the derivative of the curve evaluated at the point where the curve and tangent line meet.

at {{{x= 2}}} , {{{y = -2^2- 3*2 + 4=-4-6+4=-6}}}

point of tangency is at ({{{2}}},{{{-6}}})

{{{y}}}’ ={{{ -2x- 3}}}

compute the value of the derivative function at {{{x=2}}}

{{{y}}}'{{{(2)=-2*2-3=-7}}}

so, slope is {{{m=-7}}}

use the point-slope formula of a line, substitute the values above

{{{y-y1=m(x-x1)}}}

{{{y-(-6)=-7(x-2)}}}

{{{y+6=-7x+14}}}

{{{y=-7x+14-6}}}

{{{y=-7x+8}}}=> tangent line


{{{ drawing( 600, 600, -10, 10, -10, 10,circle(2,-6,.12), locate(2,-6,p(2,-6)),
graph( 600, 600, -10, 10, -10, 10, -x^2- 3x + 4, -7x+8)) }}}