y= ax^2+bx-4. How would i find the equation of this parabola with the points (0,-4) (1,0) (2,2) (4,0) and 6,-10)?

Answer 1

Use the first few given points
for (0,-4) : -4 = 0 + 0 - 4 , nothing new learned here
for (1,0) : 0 = a + b - 4 ----> a+b+4 or b = 4-a
for (2,2) : 2 = 4a + 2b - 4 --- 2a + b = 3

use substitution
2a + (4-a) = 3
a = -1
then b = 4-(=1) = 5

y = -x^2 + 5x - 4
test if the other points satisfy, they do!

Answer 2

You must write equation for any pair of points.

y=ax^2+bx-4

x=0 y= -4

-4=a*0^2+b*0+c

0+0+c= -4

c= -4

x=1 y=0

0=a*1^2+b*1-4

a*1+b*1-4=0

a+b-4=0

a=4-b

x=2 y=2

y=ax^2+bx-4

2=(4-b)*2^2+b*2-4

(4-b)*4+2b-4=2

16-4b+2b=2+4

-2b=6-16

-2b= -1 Divide with -2

b=5

a=4-b

a=4-5

a= -1

So:

a= -1 b=5 c= -4

y=ax^2+bx+c= -x^2+5x-4

y=-x^2+5x-4

Checking of results:

For x=4

y=ax^2+bx-4

y= -4^2+5*4-4

y= -16+20-4

y=4-4

y=0

Correct value

For x=6

y=ax^2+bx-4

y= -6^2+6*5-4

y= -36+30-4

y= -6-4

y= -10

Correct value

Answer 3

I make one mistake in typig.

-2b= -1 Divide with -2 is mistake

-2b= -10 Divide with -2 is correct