Please help I'm giving 11 pt

These are all in vertex form,

y = a(x - p)² + q

When a is positive this makes a CUP (concave up positive) and has a minimum, not a maximum.  When a is negative then the maximum is just when the squared term is 0, in which case

y = q

is the maximum.

From the figure g(x) has a maximum of 3 at x=2.

A.  Positive coefficient, CUP, no max

B. CUP, no max

C. Negative coefficient, will have a max, at y=3, not greater than 3.

D. Negative coefficient, max at y=4

D

geometry: triangles

solvers

lessons

answers archive

call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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