Respuesta :
The standard form equation of a prabola is
[tex]y=ax^2+bx+c\text{.}[/tex]We are given 3 points of the parabola; (3,-20), (5,0) and (8,0). We'll substitute these values in the standard form equation, which will give us a system of 3 equations we can use to find a, b and c.
From (3,-20) we have:
[tex]-20=9a+3b+c,[/tex][tex]9a+3b+c=-20.[/tex]From (5,0):
[tex]0=25a+5b+c,[/tex][tex]25a+5b+c=0.[/tex]From (8,0):
[tex]64a+8b+c=0.[/tex]So our system of equations is:
[tex]9a+3b+c=-20,[/tex][tex]25a+5b+c=0,[/tex][tex]64a+8b+c=0.[/tex]Let's subtract the second equation from the third:
[tex]39a+3b=0.[/tex]Now let's subtract the first from the second:
[tex]16a+2b=20.[/tex]We know have a system of two equations:
[tex]39a+3b=0,[/tex][tex]16a+2b=20.[/tex]Let's multiply the first by 2 and the second by 3:
[tex]78a+6b=0,[/tex][tex]48a+6b=60.[/tex]Let's subtract the second from the first:
[tex]30a=-60,[/tex]From which, by dividing both sides by 30 we get
[tex]a=-2.[/tex]Using this value on either of the equations of the system of two equations will give us the value of b:
[tex]16(-2)+2b=20,[/tex][tex]-32+2b=20,[/tex][tex]2b=52,[/tex][tex]b=26.[/tex]Using the values of a and b in any of the equations of the system of three equations will give us the value of c:
[tex]25(-2)+5(26)+c=0,[/tex][tex]-50+130+c=0,[/tex][tex]80+c=0,[/tex][tex]c=-80.[/tex]And so, the equation of the parabola in standard from that contains all three given points is:
[tex]y=-2x^2+26x-80.[/tex]
