Linear systems from positions a quadratic passes through? I don't understand the question.
$$y = ax^2 + bx + c$$
passes through the points $(1,-4),(-1,0),(2,3)$.
Write down a linear system (unknowns $a,b,c$) of three equations relating the unknowns to each of the points that the equation passes through?
I've been trying to solve this but i don't really understand the question.
The part; 'relating the unknowns to each of the points' i don't grasp.
Thanks for all advice/answers.
(i realise that this is very simple to a lot of you so sorry in advance)
 A: Remember, every single point on any graph MUST satisfy the equation of the graph. For example, $(1, -1)$ is on the graph of $y=x-2$ because $-1=1-2$. In your question, three points are given. These three points must satisfy the equation $y=ax^2+bx+c$. Each point is in the form $(x,y)$. So, to find the equation of the graph, we plug in the points into the equation. This will give us three equations with three unknown variables, creating a system of equations, which is what you wanted.
For $(1, -4)$:
$$-4=a(1)^2+b(1)+c$$
$$-4=a+b+c$$
For $(-1,0)$:
$$0=a(-1)^2+b(-1)+c$$
$$0=a-b+c$$
For $(2,3)$:
$$3=a(2)^2+b(2)+c$$
$$3=4a+2b+c$$
So, the linear system you were looking for is:
$$-4=a+b+c$$
$$0=a-b+c$$
$$3=4a+2b+c$$
To make it look more nice, I will put the numbers on the right and the unknowns on the left.
$$a+b+c=-4$$
$$a-b+c=0$$
$$4a+2b+c=3$$
A: Hint For the point $(2,3)$ you get the equation $3=4a+2b+c$. Do the same for the other $2$ points and you have the system.
A: Simply substitute each point into the equation. You will get a linear system of three equations:
$$a + b + c = -4\\
a - b + c = 0\\
4a + 2b + c = 3$$
I believe that by substituting the points in, you are somewhat relating the unknowns to each of the points.
A: We know that the point $(1,-4),(-1,0), (2,3)$ lie on the curve $y=ax^2 + bx +c$. This implies that we must have
$a+b+c=-4$
$a-b+c=0$
$4a+2b+c=3$
And I suppose this is what the question is asking for.
A: Another way:
Write $$y=ax^2+bx+c=A(x-1)(x+1)+B(x+1)(x-2)+C(x-2)(x-1)$$
At $x=1,y=-4\implies -4=B(1+1)(1-2)=-2B\iff B=2$
