Preamble It's not linear. A function of the form y = ax + b does not give the same y value for 2 different x values unless it is a line of the form y = a. That would mean that when x is 0, the y value should be (in this case) - 10. It is 8. So this function is not linear.
Could it be a quadratic? Maybe. The y intercept is at y = -8 Since it is a point, the y intercept is (0,-8).
So let's keep looking. There's a root between x = 1 and x = 2. How do you know? The sign changes. Is that root possible? again maybe.
If this is a quadratic, it's formula will be y = ax^2 + bx + c The y intercept is c. That means when x = 0 y = c y = ax^2 + bx - 8 Now use any two points to solve for a and b
But before we do that, we should take a look at the points more closely. The only way that - 10 could be the same y value for 2 different x values is if there is a local minimum (in this case) between them.
We know it has to be a minimum because -8 (the y part of the y intercept) is above - 10. and the value of the quadratic is above - 10
for (1,-4) we get y = ax^2 + bx - 8 -4 = a(1)^2 + b(1) - 8 -4 = a + b - 8 Add 8 to both sides. 4 = a + b
For (2,2) we get y = ax^2 + bx - 8 2 = a(2)^2 + b(2) - 8 2 = 4a + 2b - 8 10 = 4a + 2b Divide through by 2 5 = 2a + b b = 5 - 2a Put this in the equation above. a + b = 4 a + 5 - 2a = 4 -a = -1 a = 1
5 = 2a + b 5 = 2(1) + b b = 3
So the formula according to this is y = x^2 +3x - 8 Is this possible? If it is, then (-2,-10) should be a point on the quadratic graph. Is it?
y = x^2 + 3x - 8 -10 =? (-2)^2 + 3(-2) - 8 Does the right side give - 10 -10 =? 4 - 6 - 8 -10 =? - 10
It works. There is one last thing we should check. What are the roots. a = 1 b = 3 c = - 8
Using the quadratic formula (and I will leave this to you). one root is (1.70,0) and the other one is (-4.7,0)
The first root confirms what we said about a root being between x = 1 and x = 2.
Conclusion The points are on a quadratic. One last thing, I've included a graph just to confirm everything I've written.