How do you find the equation of the tangent line to the curve # -x^2+5x# at (-1,-5)?

Set #color(white)("d")y=-x^2+5x" ".................Equation(1)#

Check: Set #x=-1 => y= -1+5(-1)= -6#

#color(red)("The point (-1,-5) is NOT on the curve "y=-x^2+5)#
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#color(magenta)("Method assuming that (-1,-5) is on the curve")#

Increment #x# by the minute amount of #deltax#

As you have changed the #x# part in doing this then the #y# part will also change.

Let the resuting change in #y# be #deltay# so after the change we have:

#y+deltay=-(x+deltax)^2+5(x+deltax)" "...Equation(2)#

#y+deltay=-(x^2+2xdeltax+(deltax)^2)+5x+5deltax#

#y+deltay=-x^2-2xdeltax-(deltax)^2+5x+5deltax ....Eqn(2_a)#

Apply the subtraction #Eqn(2_a)-Eqn(1)#

#y+deltay=-x^2-2xdeltax-(deltax)^2+5x+5deltax#
#ul(ycolor(white)("dddd") = -x^2color(white)("ddddddddddd.d")+5x larr" Subtract")#
#color(white)("ddd") deltay=color(white)("dd")0x^2-2xdeltax-(deltax)^2+0x+5deltax#

Set gradient as #("change in " y)/("change in " x)->(deltay)/(deltax) #

so divide both sides by #deltax# to get gradient giving:

#(deltay)/(deltax)=-(2xdeltax)/(deltax) -(deltax)^2/(deltax)+(5deltax)/(deltax)#

#(deltay)/(deltax)=-2x-deltax+5#

#lim_(deltax->0)(deltay)/(deltax)=-2x-lim_(deltax->0)deltax+5#

#color(red)(f'(x)->dy/dx=-2x-0+5 larr" Gradient")#
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Thus for the straight line #color(blue)(y=color(red)(m)x+c)# we have #color(red)(m=-2x+5)#

It is given that the point is #P_1(x,y)=(-1,-5)#

Thus substitute #-1# for #x => m=-2(-1)+5=+7->+7/1=("change in y")/("change in x")#

So #y=mx+c color(white)("d")->color(white)("d")y=7x+c#

Substitute in the known point to determin #C#

#y=7x+c color(white)("dddd")->color(white)("dddd") -5=7(-1)+c#

#color(white)("ddddddddddddd")->color(white)("dddd")-5+7=c#

#color(white)("ddddddddddddd")->color(white)("dddddddd.d")c=2#

So the equation of the tangent at #P_1(x=-1,y=-5)" is "#

#y=7x+2#

#y=-x^2+5x#

thus gradient is #-2x+5#

Set the known point #P_1->(x,y)=(-1,y)#

Thus we have:

#y=-(-1)^2+5(-1) = -6#

Setting #P_1->(x,y)=(-1,-6)#
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Gradient at #x=1 -> m=-2(-1)+5 = +7# giving:

#y=7x+c#

Known point #(x,y)->-6=7(-1)+c color(white)("d")=>color(white)("d") c=+1#

Thus the equation of the tangent is: #y=7x+1#