Respuesta :

Given points (-4, -113) (5,-293) and (-1,7)

We frame 3 equations using the given points

[tex]y=ax^2+bx+c[/tex]

Plug in (-4, -113)

[tex]-113=a(-4)^2+b(-4)+c[/tex]

-113 = 16a -4b + c -------> first equation

Plug in (5,-293)

[tex]-293=a(5)^2+b(5)+c[/tex]

-293 = 25a +5b + c----> second equation

Plug in  (-1,7)

[tex]7=a(-1)^2+b(-1)+c[/tex]

7 = 1a -1b + c -----> third equation

Now we use the three equation and solve for a,b,c

Use first and second equation and subtract it

-113 = 16a -4b + c

+293 = -25a -5b - c

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180 = -9a + -9b (divide the whole equation by 9)

20= -1a -1b --------------> fourth equation

Subtract third equation from first equation

-113 = 16a -4b + c

-7 = -1a +1b - c  

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-120 = 15a - 3b (divide by -3 on both sides)

40 = -5a +b --------------> fifth equation

Use fourth and fifth equation

20= -1a -1b

40 = -5a +b

Add both equations

60 = -6a

so a= -10

Now we use the fourth equation

20= -1a -1b

20 = -1(-10) -1b

20 = 10 - 1b

-1b = 10

so b =-10

Now plug in the values in third equation

7 = 1a -1b + c

7 = 1(-10) -1(-10) + c

7 = c

We got a=-10, b=-10  and c=7

So equation becomes [tex]y=-10x^2-10x+7[/tex]