algebra 2 unit 2 question. connexus.

From the graph we can infer that our function crosses the x-axis at -2.5, 0.5, and 1.5, so those are the zeroes of the function.
Next, lets convert those decimals into fractions by adding a denominator 1 to each decimal and multiply both its numerator and denominator by a power 10 that lets us get rid of the decimal point:
[tex]( \frac{-2.5}{1} )( \frac{10}{10} )= \frac{-25}{10} =- \frac{5}{2} [/tex]
[tex]( \frac{0.5}{1} )( \frac{10}{10} )= \frac{5}{10} = \frac{1}{2} [/tex]
[tex]( \frac{1.5}{1} )( \frac{10}{10} )= \frac{15}{10} = \frac{3}{2} [/tex]

Since we know that each one of those points is a zero of the function ([tex]x=0[/tex]), we can replace each value into our [tex]x=0[/tex], and work backwards to get our binomials:
[tex]x=- \frac{5}{2} [/tex]
[tex]2x=-5[/tex]
[tex]2x+5=0[/tex]
So, our first binomial is [tex](2x+5)[/tex]

[tex]x= \frac{1}{2} [/tex]
[tex]2x=1[/tex]
[tex]2x-1=0[/tex]
So, our second binomial is [tex](2x-1)[/tex]

[tex]x= \frac{3}{2} [/tex]
[tex]2x=3[/tex]
[tex]2x-3=0[/tex]
So, our next binomial is [tex](2x-3)[/tex]

Now lets put all together to find the equation of our function:
[tex]y=(2x+5)(2x-1)(2x-3)[/tex]
And of course, we can rewrite it in a different order:
[tex]y=(2x-1)(2x-3)(2x+5)[/tex]

We can conclude that the equation that most likely represents the graph of the function is [tex]y=(2x-1)(2x-3)(2x+5)[/tex]