Question

The parabola with equation $y=ax^2+bx+c$ is graphed below:

(attached)

The zeros of the quadratic $ax^2 + bx + c$ are at $x=m$ and $x=n$, where $m>n$. What is $m-n$?

Answer 1

2

Explanation:

Source: AOPS

Answer 2

`m-n=2`

Explanation:

Instead of using the standard form, we can use the vertex form of a quadratic equation:

`f(x)=a(x-h)^2+k`

Where a is the leading coefficient, and (h, k) is our vertex.

Our vertex point is at (2, -4). So, let’s substitute 2 for h and -4 for k:

`f(x)=a(x-2)^2-4`

Now, we need to determine a.

We know that it passes through the point (4, 12). So, when x is 4, y must be 12. In other words:

`12=a((4)-2)^2-4`

Solve for a. Subtract within the parentheses:

`12=a(2)^2-4`

Add 4 to both sides:

`16=a(2)^2`

Square:

`16=4a`

Solve:

`a=4`

Thererfore, the value of a is 4.

So, our function is:

`f(x)=4(x-2)^2-4`

Now, let’s find our roots. Set the equation to 0 and solve for x:

`0=4(x-2)^2-4`

`4=4(x-2)^2\\1=(x-2)^2\\x-2=\pm1 \\ x=2\pm1 \\ x=3\text{ or } 1`

So, our roots are 1 and 3.

The greater root is 3 and the lesser root is 1.

Therefore, m-n, where m>n, is 3-1 or 2.

Our final answer is 2.