# Algebra

Given the following three points, find by hand the quadratic function they represent.
(−1,−8), (0,−1),(1,2)

A. f(x)=−3x2+4x−1

B. f(x)=−2x2+5x−1

C. f(x)=−3x2+10x−1

D. f(x)=−5x2+8x−1

Determine if the following set of ordered pairs represents a quadratic function. Explain.

(5, 7), (7, 11), (9, 14), (11, 18)

A. The y-values go up by the square of the x-value (22=4). Therefore, the ordered pairs represent a quadratic equation.

B. The y-values go up by the square of the x-value (22=4). Therefore, the ordered pairs do not represent a quadratic equation.

C. Since the differences between the x-values is 2 and the differences between the y-values is 4, that means that the differences between the differences of the y-values are all zero. Therefore, the ordered pairs represent a quadratic equation.

D. Since the differences between the differences of the y-values is not consistent, the ordered pairs do not represent a quadratic equation.

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1. let the function be y = ax^2 + bx + c
for (0,-1) , -1 = 0 + 0 + c , so c = -1
and our equation is y = ax^2 +bx - 1
for (-1,-8)
-8 = a - b - 1 ----> a - b = -7
for (1,2)
2 = a + b - 1 ---> a + b = 3
2a = -4
a = -2
in a+b=3 , -2+b = 3
b = 5

so which choice matches my answer?

do the points
, (11, 18) represent a function?
I don't know what properties you learned, but I would do it this way,
x -- y --1st diff -- 2nd diff
5 7
7 11 -- 4
9 14 -- 3 --- -1
11 18 -- 4 --- +1

The second differences should be a constant, they are not, so .... no quadratic

another way would be to find the quadratic using the first 3 points, then testing
if the 4th given point satisfies this equation, it does not, so NO quadratic

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Reiny
2. first one only, you do the rest
note the middle point (0,-1) works for all the equations, y = -1 for any x
now the first point (-1 , -8)
A , -3 - 4 - 1 = -8 ok so far
B , -2 - 5 - 1 = -8 ok so far
C , -3 -10 -1 =-14 no way
D , -5 - 8 - 1 = -14 no way
so it is A or B
try last point (1,2)
A gives -3 + 4 - 1 = 0
B gives -2 + 5 -1 = 2 !!!!! yes, it is B

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Damon

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