Answer to Question #147234 in General Chemistry for arjj

Question

Write an equation of a parabola that satisfies each set of conditions.
a) opens upward, congruent to y 5 x 2, and vertex (0, 4)
b) opens upward, congruent to y 5 x 2, and vertex (5, 0)
c) opens downward, congruent to y 5 x 2, and vertex (5, 0)
d) opens upward, narrower than y 5 x 2, and vertex (2, 0)
e) opens downward, wider than y 5 x2, and vertex (22, 0)
f) opens upward, wider than y 5 x 2, and vertex (1, 0)

Answer 1

Given

(1) y = 1/2*x^2 whose vertex is at (0,0).

To shift (1) down by 4 units, subtract 4 from (1) and get

(2) y = 1/2*x^2 - 4

To shift (2) to the right by 4 units subtract (not add) 4 units from x in (2) and get

(3) y = 1/2*(x-4)^2 - 4

Now FOIL (x-4)^2 and (3) becomes

(4) y = 1/2*(x^2 - 8*x +16) -4 or

(5) y = 1/2*x^2 - 4*x +8 -4 or

(6) y = 1/2*x^2 - 4*x + 4

Let's check this to see if the point (4,-4) is satisfied.

Is (-4 = 1/2*4^2 - 4*4 + 4)?

Is (-4 = 8 - 16 + 4)?

Is (-4 = -8 + 4}?

Is (-4 = -4)? Yes

Answer: The equation of the parabola that has its vertex at (4.-4) and opens upward with a compression factor of 1/2 is

y = 1/2*x^2 - 4*x +4.