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9. the equation of the parabola is given
y = -1/4x^2 + 4x - 19
what are the coordinates of the vertex of the parabola ?

complette the square to get vertex form or y=a(x-h)^2+k
(h,k) is vertex
1. group x terms, so for y=ax^2+bx+c, do y=(ax^2+bx)+c 2, factor out the leading coefinet (constant in front of the x^2 term), basicallly factor out a 3. take 1/2 of the linear coefient (number in front of the x), and square it ,then add negative and positive of it inside parnthases 4. complete the squre and expand

so
y=-1/4x^2+4x-19
group
y=(-1/4x^2+4x)-19
undistribute -1/4
y=-1/4(x^2-16x)-19
take 1/2 of -16 and squer it to get 64 then add neg and pos inside
y=-1/4(x^2-16x+64-64)-19
factorperfect square
y=-1/4((x-8)^2-64)-19
expand
y=-1/4(x-8)^2+16-19
y=-1/4(x-8)^2-3
vertex is (8,-3)

The answer is