\(\text{The parabola with equation $y=ax^2+bx+c$ is graphed below:}\)
\(\text{The zeros of the quadratic $ax^2 + bx + c$ are at $x=m$ and $x=n$, where $m>n$. What is $m-n$? }\)
The x coordinate of the vertex can be found as -b / (2a) = 2 ⇒ -b = 4a ⇒ b = -4a
So.....we have the following quadratic
y = ax^2 - 4ax + c
And since (2,1) and (-4, -3) are on the graph, we have that
a(2)^2 - 4a(2) + c = 1 ⇒ 4a - 8a + c = 1 ⇒ -4a + c = 1 ⇒ -32a + 8c = 8 (1)
a(-4)^2 - 4a(-4) + c = -3 ⇒ 16a + 16a + c = -3 ⇒ 32a + c = -3 (2)
Add (1) and (2) and we have that
9c = 5
c = 5/9
And
-4a + 5/9 = 1
-4a = 4/9
a = -1//9
And b = -4(-1/9) = 4/9
So....the quadratic is
y = (-1/9)x^2 + (4/9)x + 5/9
Setting this to 0 and we have that
(-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 set each factor to 0 and solve for x and we have that
x - 5 = 0 x + 1 = 0
x = 5 = m x = -1 = n
So....m - n = 5 -(-1) = 6