# B Quadratic term's relationships

1. Mar 29, 2016

### thetexan

I know everyone here must know this but it is a new revelation to me and very interesting. But I have a question about the reason behind the following...

Take $x^2+10x+10$ for example.

Any coefficient of $x^2$ modifies the "zoom" of the parabola, for lack of a better word, and determines whether it is a "up" parabola or "down" parabola.

The x term and it's coefficient determines the center of the parabola. The center is always 1/2 of the inverse of the coefficient. For example 10x results in a centerline of -5.

The constant gives the y intercept.

The x2 coefficient is related to the x coefficient in this respect...the parabola centerline at x will always be 1/2 of inverse of the x coefficient divided by the coefficient of the $x^2$ coefficient!

For example...

$2x^2 + 60x + 20$ results in....

1. a y intercept of 20
2. a parabola centerline of $( (+60/2)*-1) / 2$ or -15

I'm trying to understand why this should be. Why or how does the $x^2$ coefficient of 2 have any relationship with the x coefficient of 60 such that the two together determine the parabola centerline?

Any insights would be appreciated.

Thanks,
tex

2. Mar 29, 2016

### blue_leaf77

$$y=2x^2 + 60x + 20 = 2(x^2+30x+225)+20-450 = 2(x+15)^2-430$$
This is a parabola resulting from a translation of a centered parabola $y=2x^2$ to the left by 15 and then down by 430.

3. Mar 29, 2016

### PeroK

Perhaps most simply:

$2x^2 + 60x + 20 = 2(x^2 +30x + 10)$

This graph on the left is twice the height of the graph on the right at every point. So, it has a different shape, twice the y intercept but the same centre line.

I mean of course twice the function in the brackets on the right!

4. Mar 30, 2016

### symbolipoint

Understanding the general form and the standard form for the equation of a parabola is very important and useful. The process of Completing the Square is how you change from the general form to the standard form. The standard form equation is easier to graph.

5. Mar 30, 2016

### robphy

Here's a kinematic way to think about things:

$y=\frac{1}{2}a_0t^2+v_0t+y_0,$
where $a_0$ is the [initial and constant] acceleration, $v_0$ is the initial velocity ($v$ when $t=0$), and $y_0$ is the initial position.

On a position-vs-time graph,
$y_0$ is the y-intercept (where $t=0$)
$v_0$ is the slope at the y-intercept (based on the derivative taken below).

The centerline occurs when the slope of the graph is zero.
Physically, that is when the velocity is zero.
By taking the derivative with respect to time, we obtain the velocity function
$v\equiv\frac{dy}{dt}=a_0t+v_0$
and asking for the value of $t$ that makes this zero is: $t_{when\ v=0}=-\frac{v_0}{a_0}$.

Based on your example, $y=2t^2 + 60t + 20$
$a_0=4$ and $v_0=60$
so $t_{when\ v=0}=-\frac{60}{4}=-15$.

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