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Question

6) A parabola with the queation y = ax^2 + bx + c, with the points (-1,-10), (2,-1) and, (3,-6), find; a,b,c

Answer 1

y = ax^2 + bx + c

Plugging in (-1, -10) for (x, y) gives us

-10 = a(-1)^2 + b(-1) + c
-10 = a - b + c

Plugging in (2, -1) for (x, y) gives us

-1 = a(2)^2 + b(2) + c
-1 = 4a + 2b + c

Plugging in (3, -6) for (x, y) gives us

-6 = a(3)^2 + b(3) + c
-6 = 9a + 3b + c

Three equations, three unknowns:

a - b + c = -10
4a + 2b + c = -1
9a + 3b + c = -6

From the first equation, a - b + c = -10, so
a = b - c - 10

Plugging this into the second and third equations,

4a + 2b + c = -1
4(b - c - 10) + 2b + c = -1
4b - 4c - 40 + 2b + c = -1
6b - 3c = -1

9a + 3b + c = -6
9(b - c - 10) + 3b + c = -6
9b - 9c - 90 + 3b + c = -6
6b - 8c = 84
3b - 4c = 42

Now we have two equations, two unknowns.
6b - 3c = -1
3b - 4c = 42

This time let's use elimination to solve this. Multiplying the second equation by (-2) yields
6b - 3c = -1
-6b + 8c = -84

Adding the equations together,
5c = -85, so
c = -17

Since c = -17, we can now get b:
6b - 3c = -1
6b - 3(-17) = -1
6b + 51 = -1
6b = -52
b = -26/3

b = -26/3, c = -17

Now, we can easily get a.

a - b + c = -10
a - (-26/3) + (-17) = -10
a + 26/3 = 7
a = 7 - 26/3
3a = 21 - 26
3a = -5
a = (-5/3)

So a = (-5/3), b = (-26/3), c = -17, making the parabola

y = (-5/3)x^2 - (26/3)x - 17

Test:

For x = -1, y = (-5/3) + 26/3 - 17 = 21/3 - 17 = 7 - 17 = -10.
So the curve goes through (-1, -10).

I made an error somewhere. But those are the basic steps.

Answer 2

1) -10 = a(-1)^2 +b(-1) + c
2) -1 = a(2)^2 + b(2) + c
3) -6 = a(3)^2 + b(3) + c

-10 = a - b + c
-1 = 4a + 2b + c
-6 = 9a +3b +c

3-2)
-5 = 5a + b

3-1)
4 = 8a + 4b
1 = 2a + b

(3-1)-(3-2))
6 = -3a
a = -2
b = 5
c = -3

check
-2 -5 -3 = -10
-8+10 - 3 = -1
-18 + 15 - 3 = -6
y=-2x^2 +5x -3

Answer 3

You have three equations in three unknowns.

y = ax² + bx + c

Plug in the three points.

-10 = a - b + c
-1 = 4a + 2b + c
-6 = 9a + 3b + c

Solving we get

a = -2
b = 5
c = -3