Linear systems from positions a quadratic passes through?

Question

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)

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

Another way:

Write $$y=ax^2+bx+c=A(x-1)(x+1)+B(x+1)(x-2)+C(x-2)(x-1)$$

At $x=1,y=-4\implies -4=B(1+1)(1-2)=-2B\iff B=2$

Answer 5

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. $$a+b+c=-4$$ $$a-b+c=0$$ $$4a+2b+c=3$$