the solution to nx^2+PX+Q=0 are 4 and -2/5. what information does this tell you about the graph of f(x)=nx^2+px+q?

3 years ago

Answered By Abdullah A

The solutions or roots to a quadratic equation tell you where the graph crosses the x axis. Therefore, the graph crosses the x-axis 2 times at x=4 and x= -2/5

3 years ago

Answered By Emily D

Abdullah gave a great short answer; I'm just elaborating in case you would like more context.

We're told that when x = 4 or x = -2/5, f(x) is 0

N*4^{2} + P*4 + Q = 0

and

N*(-2/5)^{2} + P*(-2/5) + Q = 0

This means our graph is crossing the x-axis (the horizontal line) when x = 4 or x = -2/5. This would look similar to one of the two graphs I have drawn below. It won't look exactly the same, but it definitely crosses over the x-axis in two unique spots.

With your equation, the two cross-overs happen at x = -2/5 and x = 4. These are called the "roots" of the equation. In other words: the roots of your equation tell you where the function crosses the x-axis on a graph.

3 years ago

## Answered By Abdullah A

The solutions or roots to a quadratic equation tell you where the graph crosses the x axis. Therefore, the graph crosses the x-axis 2 times at x=4 and x= -2/5

3 years ago

## Answered By Emily D

Abdullah gave a great short answer; I'm just elaborating in case you would like more context.

We're told that when x = 4 or x = -2/5, f(x) is 0

N*4

^{2}+ P*4 + Q = 0and

N*(-2/5)

^{2}+ P*(-2/5) + Q = 0This means our graph is crossing the x-axis (the horizontal line) when x = 4 or x = -2/5. This would look similar to one of the two graphs I have drawn below. It won't look exactly the same, but it definitely crosses over the x-axis in two unique spots.

With your equation, the two cross-overs happen at x = -2/5 and x = 4. These are called the "roots" of the equation. In other words: the roots of your equation tell you where the function crosses the x-axis on a graph.

## Attached Graph: