Answer:
y = (-1/3)x^2 - 2x + 5/3
Explanation:
To find the quadratic equation that passes through these points, we can start by using the standard form of a quadratic equation:
y = ax^2 + bx + c
We know that the graph passes through the point (-2,2), so we can substitute these values into the equation:
2 = a(-2)^2 + b(-2) + c
Simplifying this equation, we get:
4a - 2b + c = 2
We also know that the graph has x-intercepts at (-1,0) and (6,0). This means that when x = -1 and x = 6, the value of y (i.e., the height of the graph) is 0. We can use these two points to write two more equations:
0 = a(-1)^2 + b(-1) + c
0 = a(6)^2 + b(6) + c
Simplifying these equations, we get:
a - b + c = 0
36a + 6b + c = 0
Now we have a system of three equations:
4a - 2b + c = 2
a - b + c = 0
36a + 6b + c = 0
We can solve for a, b, and c using any method of solving systems of equations. One way is to use substitution:
From the second equation, we get:
c = b - a
Substituting this into the other two equations, we get:
4a - 2b + (b - a) = 2
36a + 6b + (b - a) = 0
Simplifying these equations, we get:
3a - b = 1
35a + 7b = -b
Multiplying the first equation by 7 and adding it to the second equation, we eliminate b:
21a = -7
a = -1/3
Substituting this value of a into the first equation, we can solve for b:
4(-1/3) - 2b + (b - (-1/3)) = 2
-4/3 - b + b + 1/3 = 2
b = -2
Finally, we can use either of the first two equations to solve for c:
c = a - b = (-1/3) - (-2) = 5/3
So the quadratic equation in standard form is:
y = (-1/3)x^2 - 2x + 5/3