Find the minmum value of x^2-4x+y^2+8y+20=?

we don’t know what it’s equal to, so we set it equal to ?, if you know that ? is, then input it

what we can do is try to convert it to a conic section

complete the squares

we don’t know what it’s equal to, so we set it equal to ?, if you know that ? is, then input it

what we can do is try to convert it to a conic section

complete the squares

(x^2-4x)+(y^2+8y)+20=?

take 1/2 of linear coefient and square it and add negative and positive insde (linear is 1st degree)

(x^2-4x+4-4)+(y^2+8x+16-16)+20=?

factor perfect squares

((x-2)^2-4)+((y+4)^2-16)+20=?

(x-2)^2-4+(y+4)^2-16+20=?

(x-2)^2+(y+4)^2=?

we see this is the equation of a circle

in form

(x-h)^2+(y-k)^2=r^2

center is (h,k)

radius is r

so the lowest point is r units down from (h,k), or the point (h,k-r)

we know that te equation is

(x-2)^2+(y-(-4))^2=?

?=r^2

√?=r

center is (2,-4)

therefor

the minimum value, where the equation is equal to ?, is (2,-4-√?)

good luck, and may the force be with you