Find Equation of Parabola, given three points

Question

(0,4) (1,3) (2,6)

I know I have to use the equation for a parabola whose axis of symmetry is vertical:

y = ax^2 +bx + c

The first step is to plug the points into an equation:

4= a(0)^2 +b(0) +c

4= a+c This is where I get confused. My book says C=4?

3= a(1)^2 + b(1) +c

3= a +b +c This is more confusing, book says a+b= -1 ??

6= a(2)^2 +b (2) +c

6= 4a+2b+c This comes to 2a +b =1 ????

Please Help!!

Thank You,

Brian

Answer 1

maybe this will help

y=ax^2+bx+c, use (0,4)

4=a(0^2)+b(0)+c

4=0+0+c

c=4

y=ax^2+bx+c, use (1,3)

3=a(1^2)+b(1)+c

3=a+b+c

y=ax^2+bx+c, use (2,6)

6=a(2^2)+b(2)+c

6=4a+2b+c

c=4, so...

3=a+b+4 and a+b=-1

6=4a+2b+4 and 4a+2b=2 and simplifying gives 2a+b=1

2a+b=1

a+b=-1

subtract

a=2 (and c=4)

using 3=a+b+c we get 3=2+b+4 and b=3-2-4, b=-3

y=ax^2+bx+c

y=2x^2-3x+4 is the equation

Answer 2

You have mistake in 1st equation. Must be c = 4

Then 2nd equation gives you a + b + 4 = 3 and a + b = - 1.

Last equation: 4a + 2b + 4 = 6, so 4a + 2b = 2 or 2a + b = 1. if we subtract 1st and 3rd equations we get a = 2

Then b = - 1 - a = - 1 - 2 = - 3

So equation of parabola: y = 2x² - 3x + 4