The equation for a parabola has the form y = ax^2 + bz + c, where a, b, and c are constants and aメ0. find an equation for the parabola that passes through the points (-1,14), (2,-7), and (5, 8) y-

a = 2, b = -9, c = 3

Step-by-step explanation:

Replacing x, y values of the points in the equation y = a*x^2 + b*x +c give the following:  

(-1,14)

14 = a*(-1)^2 + b*(-1) + c  

(2,-7)

-7 = a*2^2 + b*2 + c  

(5, 8)  

8 = a*5^2 + b*5 + c  

Rearranging:

a - b + c = 14  

4*a + 2*b + c = -7

25*a + 5*b + c = 8

This is a linear system of equations with 3 equations and 3 unknows. In matrix notation the system is A*x = b whith:

A =

1    -1  1

4    2  1

25  5  1

x =

a  

b

c

b =

14

-7

8

Solving A*x = b gives x = Inv(A)*b, where Inv(A) is the inverse matrix of A. From calculation software (I used Excel) you get:

inv(A) =  

0.055555556-0.111111111   0.055555556

-0.3888888890.444444444   -0.055555556

0.5555555560.555555556-0.111111111

inv(A)*b

2

-9

3

So, a = 2, b = -9, c = 3

  q

step-by-step explanation:

each line from a vertex to the midpoint of the opposite side is called a median. the point where the medians intersect, point q, is the centroid.

the centroid of δabc is point q.