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Trava [24]
3 years ago
14

Vertex form is f(x)=a(x-p)^2 +q. How do i determine a?

Mathematics
1 answer:
slamgirl [31]3 years ago
8 0
Y1 is the simplest parabola.  Its vertex is at (0,0) and it passes thru (2,4).  This is enough info to conclude that y1 = x^2.

y4, the lower red graph, is a bit more of a challenge.  We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).

Let's try this:  assume that the general equation for a parabola is 
y-k = a(x-h)^2, where (h,k) is the vertex.  Subst. the known values,

-3-(-4) = a(2-0)^2.  Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.

The equation of parabola y4 is   y+4 = (1/4)x^2

Or you could elim. the fraction and write the eqn as 4y+16=x^2, or

4y = x^2-16, or    y = (1/4)x - 4.  Take your pick!  Hope this helps you find "a" for the other parabolas.


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fgiga [73]

ANSWER

a_n = 11 - 3n

EXPLANATION

The explicit rule for an arithmetic sequence is given by:

a_n = a_1 + d(n - 1)

It was given that:

a_1 = 8

and

a_2 = 3

The common difference is

d = 5 - 8 =  - 3

We plug in the values into the explicit formula to get,

a_n = 8+  - 3(n - 1)

Expand to get,

a_n = 8 - 3n  + 3

The explicit rule is

a_n = 11 - 3n

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