Question

2. Your friend wants to write the equation of a quadratic function 1 5. 3.r = 1) that has zeros at 3 and 3. They want to use the binomials (3.7" – 5) and You insist that they must use the binomials 3 and (3) Who is correct? Explain.

Answer 1

Let's expand both options, your friend's and yours, as shown below

$\begin{gathered} y=(3x+1)(3x-5)=9x^2-12x-5=9(x^2-\frac{4}{3}x-\frac{5}{9}) \\ \text{and} \\ y=(x+\frac{1}{3})(x-\frac{5}{3})=x^2-\frac{4}{3}-\frac{5}{9} \end{gathered}$

Then, both equations are the same besides a constant that will not affect the zeros of the functions, as shown below

$\begin{gathered} y=(3x+1)(3x-5) \\ x=-\frac{1}{3} \\ \Rightarrow y=(-1+1)(-1-5)=0 \\ \text{and} \\ x=\frac{5}{3} \\ \Rightarrow y=(5+1)(5-5)=0 \\ \end{gathered}$

And

$\begin{gathered} y=(x-\frac{5}{3})(x+\frac{1}{3}) \\ x=-\frac{1}{3} \\ \Rightarrow y=(-\frac{1}{3}-\frac{5}{3})\cdot0=0 \\ x=\frac{5}{3} \\ \Rightarrow y=0\cdot(\frac{6}{3})=0 \end{gathered}$

Both your friend and you are correct. The functions are the same with exception of a constant that multiplies the whole function (a scale factor); despite that, the zeros are the same for both functions

$(3x-5)(3x+1)=9(x+\frac{1}{3})(x-\frac{5}{3})$