a. (-8,2); x= -8; up
b. (-8,-2); x= -8; down
c. (8,-2); x= 8; up
d. (8,2); x=8; up
2007-01-27 12:27:19 · 4 answers · asked by lana l 1 in Science & Mathematics ➔ Earth Sciences & Geology
The given equation of Parabola is
y=1/2(x-8)^2+2
Since this equation has sqaured term i.e (x-8)^2 so the value of y cannot be NEGATIVE.
also it has 2 as aditional term beside squred term so for any value of x ,y value will be SHIFTED UPWARD by 2 unit in y direction.
Hence Vertex will be having y coordinate positive.this keeps options a) and d) only.
Now x coordinates will be such that it will be at mimum distance from X axis. ie. 2 as y coordinate.
since at x=8 sqaured term will be zero so x=8 will X coordinate.
hence vertex will be (8,2) and opeening upwards.
so option (d) satisfies te gievn conditions.
2007-01-27 15:19:42 · answer #1 · answered by rajeev_iit2 3 · 0⤊ 0⤋
bear in mind that the formulation for a parabola is y=a(x-h)^2+ok the region (h,ok) is your vertex and h=axis of symmetry. on your case, a is effective so your parabola opens up. Your vertex is (-3,-5) and your axis of symmetry at x=-3.
2016-11-27 23:22:17 · answer #2 · answered by ? 4 · 0⤊ 0⤋
y=1/2 (x^2 - 16x + 64) + 2 = 1/2x^2 -8x +34
we have to find the derivative
y' = 2 * 1/2x -8 = 0
x=8
y = 1/2 * 64 - 8*8 + 34 = 2
The answer is d)
2007-01-27 12:38:31 · answer #3 · answered by really? 2 · 0⤊ 0⤋
e. all of the above
2007-01-27 12:35:07 · answer #4 · answered by Anonymous · 0⤊ 0⤋