Here's another way
In the form ax^2 + bx + c, the equation of the x coordinate of the vertex is given by -b/ 2a
So..... 2 = -b / [2a] ⇒ 4a = - b ⇒ b = - 4a
And we know that
1 = a(2)^2 - 4a(2) + c and -3 = a(-4)^2 - 4a(-4) + c
1 = 4a - 8a + c -3 = 16a + 16a + c
1 = -4a + c -3 = 32a + c ⇒ 3 = -32a - c
Add the last two equations and we have that
4 = -36a
-4/36 = a
-1/9 = a
And b = -4a = -4(-1/9) = 4/9
To find c, we have
1 = (-1/9)(2)^2 + (4/9)(2) + c
1 = (-4/9) + 8/9 + c
1 = 4/9 + c
5/9 = c
So.....the function is
y = (-1/9)x^2 + (4/9)x + 5/9
To find tthe zeroes, set this = 0
(-1/9)x^2 + (4/9)x + 5/9 = 0 multiply through by - 9
x^2 - 4x - 5 = 0 factor
(x - 5) ( x + 1) = 0
Setting each factor to 0 and solving for x and we get that x = - 1 or x = 5
So m = 5 and n = -1
So
m - n = 5 - (-1) = 6
