To write a quadratic function in standard form that passes through the points (-8,0), (-2,0), and (-6,8), we need to use the fact that any quadratic function can be written in the form:
y = a x^2 + b x + c
where a, b, and c are constants. We can find these constants by substituting the coordinates of the three points into this equation and solving for a, b, and c.
Substituting the point (-8,0), we get:
0 = a (-8)^2 + b (-8) + c
Simplifying this equation, we get:
64 a - 8 b + c = 0 (Equation 1)
Substituting the point (-2,0), we get:
0 = a (-2)^2 + b (-2) + c
Simplifying this equation, we get:
4 a - 2 b + c = 0 (Equation 2)
Substituting the point (-6,8), we get:
8 = a (-6)^2 + b (-6) + c
Simplifying this equation, we get:
36 a - 6 b + c = 8 (Equation 3)
Now we have three equations (Equations 1, 2, and 3) with three unknowns (a, b, and c). We can solve this system of equations by eliminating one variable at a time. First, we can eliminate c by subtracting Equation 2 from Equation 1:
60 a - 6 b = 0 (Equation 4)
Next, we can eliminate b by subtracting Equation 2 from Equation 3:
32 a + 4 b = 8 (Equation 5)
Simplifying Equation 4, we get:
10 a - b = 0
or
b = 10 a (Equation 6)
Substituting Equation 6 into Equation 5, we get:
32 a + 4 (10 a) = 8
Simplifying this equation, we get:
72 a = 8
or
a = 1/9
Substituting this value of a into Equation 6, we get:
b = 10 (1/9)
or
b = 10/9
Finally, substituting these values of a and b into Equation 1, we get:
64 (1/9) - 8 (10/9) + c = 0
Simplifying this equation, we get:
c = 176/9
Therefore, the quadratic function that passes through the points (-8,0), (-2,0), and (-6,8) is:
y = (1/9) x^2 + (10/9) x + (176/9)
This is the standard form of the quadratic function.