Question

Find the coordinates of the point Q on the curve -y=x^2-4x+7 at which the tangent is parallel to the line y+3x=9.

Answer 1

yes EXP, anything to say ?

Answer 2

-y=x^2-4x+7 or this is
y=-x^2+4x-7 and this line is parallel to y+3x=9.or
y=-3x+9.your slope here is m=-3
also the slope of y=-x^2+4x-7 is
dy/dx=m=-2x-4 since the line is parallel therefore their slope are equal
so m=m=-2x+4=-3
-2x=-7
x=7/2 sub this to the line eq to find y
y=-x^2+4x-7
y=-(7/2)^2+4(7/2)-7 thus
y=-21/4 now you coordinate Q(x,y)=Q(7/2,-21/4) ..ans...

Answer 3

oopps typo that was
dy/dx=m=-2x+4

Answer 4

any question let me know

Answer 5

Thank you so much. (: