Find the point on the parabole y=9-x^2 closest to the point (3,9)

(x,y)=(−1,1) is the closest point on y=x2 to (−3,0)

The distance from any point (x,y) to a point (ˆx,ˆy) is
XXX√(x−ˆx)2+(y−ˆy)2

For points on y=x2 this becomes
XXXd(x)=√(x−ˆx)2+(x2−ˆy)2
and
more specifically for the point (ˆx,ˆy)=(−3,0) this becomes
XXX√(x+3)2+(x2−0)2

XX=√x4+x2+6x+9

The problem is to minimize d(x)
or equivalently (but slightly simpler) to minimize
XXXf(x)=x4+x2+6x+9

The minimum occurs when f'(x)=0
That is when
XXX4x3+2x+6=0

An obvious (by inspection) root is x=−1
(and in fact there are no other real roots)

If x=−1
then y=x2=(−1)2=1

i hope you understand :)

answer

a,b, and d

explanation

a reflection angle is more than

but less than;

an angle that measures

is a reflex angle.(true)

an angle that measures less than

is acute.(true)

provided we are considering angles greater or equal to zero.

an acute angle lies in the range.

an obtuse angle is less than 180° but more than 90°

an angle that measures 300° is obtuse (false)

an angle that measures 65° is an acute angle.(true)