# Changing Distance Between Zeroes of Parabola at 10 Units Apart

• JoeTrumpet
In summary, the parabola with equation y= x^2 + 12x + 36 is moving downward at a rate of 2 units per second. The distance between the zeros of the parabola can be found using the equation D= 2sqrt{2t}, where t is in seconds. When the zeros are 10 units apart, the rate at which they are moving apart is 2/5 units per second.
JoeTrumpet

## Homework Statement

Let g(x) = x^2 + 12x + 36, a parabola with one zero at x = -6. The parabola moves downward at a rate of 2 units/sec. How fast is the distance between the zeroes changing when they are 10 units apart?

## Homework Equations

g(x) = x^2 + 12x + 36

## The Attempt at a Solution

I figured since this is a parabola, the zeroes would simply be equidistant from -6, so at 10 units apart they would be at x=-1 and x=-11.

Next I took the derivative of y = x^2 + 12x + 36 to obtain
dy/dt = (2x + 12)(dx/dt) and I substituted -2 for dy/dt since this parabola is moving down and solved for dx/dt to get
dx/dt = -2/(2x+12). Next I substituted in -11 and -1 for x and obtained, respectively, dx/dt = 1/5 and -1/5. But this doesn't seem to make sense: doesn't this mean they're moving closer to each other, when they should be going further away? The correct answer is 2/5, so if the signs were what I expected I would've had it correct, but I can't figure out what's wrong.

It's worth mentioning that I was given a solution that was quite distinct from the manner in which I tried to solve it, but I didn't understand it either. This is the provided solution:
y = ax^2 + bx +c, therefore dc/dt = -2. Distance between roots = (-b+(b^2-4ac)^(1/2))/(2a) - (-b-(b^2-4ac)^(1/2))/(2a) = ((b^2-4ac)^(1/2))/a = (144-4c)^(1/2). Distance = 10 when c = 11, thus the rate at which they are moving apart is 2/5.

I understand it up until the last step, where I don't see how the jump was made. Thanks in advance for the help!

What do you think "the parabola is moving downward at 2 units per second" means? You can't just look at y= x2+ 12x+ 36- that's a single parabola, it isn't moving! A parabola that, at time t= 0, is y= x2+ 12x+ 36 and such that the entire parabola (i.e. y value for every x) is moving downward at 2 units per second must be y= x2+ 12x+ 36- 2t where t is in seconds. That's their "y= ax2+ bx+ c with dc/dt= -2". Here c= 36-2t and dc/dt= -2.

We can simplify that a bit by noting that x2+ 12x+ 36= (x+ 6)2 so we are really talking about y= (x+6)2- 2t. The zeros are given by (x+6)2= 2t so $x= -6\pm\sqrt{2t}$. The distance between those zeros is $D= 2\sqrt{2t}$. You could differentiate that directly but I think it is easier to write D2= 8t and then differentiate: 2D dD/dt= 8 so dD/dt= 4/D. When the zeros are 10 units apart, D= 10 so dD/dt= 4/10= 2/5 unit per second.

## 1. What is the formula for finding the distance between zeroes of a parabola at 10 units apart?

The formula for finding the distance between zeroes of a parabola at 10 units apart is d = 10√(a²+ b²), where a and b are the coefficients of the quadratic equation.

## 2. How do you determine the direction of the parabola when the distance between zeroes is 10 units?

The direction of the parabola can be determined by looking at the sign of the coefficient of the quadratic term. If the coefficient is positive, the parabola opens upwards, and if it is negative, the parabola opens downwards.

## 3. Is it possible for the distance between zeroes to be less than 10 units?

Yes, it is possible for the distance between zeroes of a parabola to be less than 10 units. This would occur when the parabola is compressed or stretched horizontally, resulting in a smaller or larger distance between zeroes, respectively.

## 4. Can the distance between zeroes of a parabola at 10 units apart be negative?

No, the distance between zeroes of a parabola cannot be negative. The distance is always measured as a positive value, regardless of the direction of the parabola.

## 5. How does changing the value of the quadratic coefficient affect the distance between zeroes at 10 units apart?

The quadratic coefficient affects the distance between zeroes by determining the steepness of the parabola. A larger coefficient will result in a steeper parabola, while a smaller coefficient will result in a flatter parabola. However, the distance between zeroes will remain at 10 units unless the horizontal stretch or compression occurs.

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