quadratic equation of the parabola that goes through the points (-2,1)(0,1)(-1,0)

Question

Write the quadratic equation of the parabola that goes through the points (-2,1) (0,1)(-1,0).

Answer 1

A quadratic equation will always have the form y = ax^2 + bx + c

Here you have three variables and three points.

Does that get you started? More of a hint after asterisks...

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Try the easiest one first, point (0, 1), sub those in for x and y

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so 1 = a(0)^2 + b(0) + c

1 = c

Now you have two more equations and two variables to find...

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Try point (-1, 0)

0 = a(-1)^2 + b(-1) + 1

0 = a - b + 1

a - b = -1

Now (-2, 1)

1 = a(-2)^2 + b(-2) + 1

0 = 4a - 2b

0 = 2a - b

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From there you can solve for a and b by any method you like. I like substitution.

a - b = -1

a = b - 1

0 = 2a - b

0 = 2(b - 1) - b

0 = 2b - 2 - b

0 = b - 2

b = 2

a - b = -1

a - 2 = -1

a = 1

Soooooo...plugging back in gives us

y = ax^2 + bx + c

y = 1x^2 + 2x + 1

y = x^2 + 2x + 1

You can check by graphing or by plugging the points back in, let's try one for fun

1 =? (-2)^2 + 2(-2) + 1

1 =? 4 - 4 + 1

1 = 1

check!

Answer 2

y = ax^2+bx+c

(-2,1) --> 1 = 4a - 2b + c

(0,1) --> 1 = c

(-1,0) --> 0 = a - b + c

substitiutes c=1:

1 = 4a-2b+1

0 = a -b + 1

subtracts 1:

0 = 4a-2b

-1 = a-b

divides first equation by 2...

solves 2nd equation for a:

0 = 2a-b

b-1 = a

solves by substitution

0 = 2(b-1)-b

0 = 2b-2-b

0 = b-2

b=2

then a = 2-1=1

y=x^2+2x+1

note that when x=0, y=1 --> (0,1)

when x=-2, y = 4-4+1=1 --> (-2,1)

when x =-1, 1-2+1=0 --> (-1,0)

so yes it checks

y=x^2+2x+1