How Do You Formulate a Quadratic Equation with Given X-Intercepts?

[unistudent123]

"Find the rule of correspondence for a quadratic equation that is the same shape as y=x^2, with the only given information being x-intercepts at (-3, 0) and (1, 0)." I cannot for the life of me figure this out, and it's making me feel extremely stupid.
I know the answer to the question is y= (x+1)^2 - 4, but I'm not getting there.

 

[tedbradly]

Best I know, I'd say the center of the function (corresponding to x = 0 for y = x^2) would be dead-center between any 2 points of the same height.

Thus, the center would be at (-3 + 1)/2 = -1

Then, I'd think the to plug in the value x = (1 - (-3))/2 = 2 to find the height of the untranslated function:

y = 2^2 = 4. Thus, wherever it is 4 originally, we want it to be zero, making the vertical translation -4.

So we add the vertical translation and plug in x minus the horizontal translation into the function:
f(x) = (x+1)^2 - 4

 

[Pengwuino]

Think of it this way, what do the x-intercepts tell you about the equation? They're the zeros of the function. So x=-3 and x=1 are the zeros.

In other words, you can factor the equation as y=(x-1)(x+3). This will expand out to a quadratic form of the equation. You can further manipulate to give the answer you have.

 

[HallsofIvy]

"Find the rule of correspondence for a quadratic equation that is the same shape as y=x^2, with the only given information being x-intercepts at (-3, 0) and (1, 0)." I cannot for the life of me figure this out, and it's making me feel extremely stupid.



I know the answer to the question is y= (x+1)^2 - 4, but I'm not getting there.

There are, in fact, an infinite number of quadratic functions that pass through (-3, 0) and (1, 0). I guess that "the same shape as y= x^2" is intended to mean that the leading coefficient is 1. Remember that you could find the intercepts, that is solve the equation x^2+ bx+ c= 0, by factoring. Work the other way.