Function Equation Question (concerning x-ints)

[Math_Junkie]

Why do the following functions yield different value even though they have the same x-intercepts?

y = (x-3)(x+1/2)^2
y = (x-3)(2x+1)^2

If a question asks you to write an equation with x-intercepts at 3 and -1/2, how do you know which equation is right?

 

[Deleted member 4993]

Why do the following functions yield different value even though they have the same x-intercepts?

y = (x-3)(x+1/2)^2
y = (x-3)(2x+1)^2

If a question asks you to write an equation with x-intercepts at 3 and -1/2, how do you know which equation is right?

Both the functions are correct answer for the question asked

 

[soroban]

Hello, Math_Junkie!

Why do the following functions yield different value even though they have the same x-intercepts?

. . \(\displaystyle \begin{array}{ccc}y &=& (x-3)\left(x+\frac{1}{2}\right)^2 \\ \\[-4mm] y &=& (x-3)(2x+1)^2 \end{array}\)

If a question asks you to write an equation with x-intercepts at 3 and -1/2,
how do you know which equation is right?

It asked you write an equation . . . any one of a zillion possible equations.


\(\displaystyle \text{While }y \:=\x - 3)\left(x + \frac{1}{2}\right)\text{ is the obvious answer (the simplest),}\)

. . \(\displaystyle \text{all of these are also correct: }\;\begin{Bmatrix} y &=& 7(x-3)\left(x+\frac{1}{2}\right) \\ \\[-3mm]y &=& (x-3)(10x+5) \\ \\[-3mm] y &=&(x-3)\left(x + \frac{1}{2}\right)(x - 2) \\ \\[-3mm] y &=& (x-3)^2\left(x+\frac{1}{2}\right)^5 \end{Bmatrix}\)