Answer:
One rational root.
B is correct.
Step-by-step explanation:
Given: The graph of [tex]f(x)=4x^3-13x^2+9x+2[/tex]
First we factor the given function.
Factor of f(x)
f(2)=0 , x-2 must be factor of f(x)
[tex]f(x)=(x-2)(4x^2-5x-1)[/tex]
Because f(2)=0
So, x=2 is rational root of f(x)
Now we find another root using quadratic formula.
[tex]ax^2+bx+c=0[/tex]
[tex]\text{Quadratic formula: }x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
where, a=4, b=-5 and c=-1
[tex]x=\dfrac{-(-5)\pm \sqrt{(-5)^2-4(4)(-1)}}{2(4)}[/tex]
[tex]x=\dfrac{5\pm\sqrt{41}}{8}[/tex]
Another roots are,
[tex]x=\dfrac{5+\sqrt{41}}{8},\dfrac{5-\sqrt{41}}{8}[/tex]
So, these are irrational root because [tex]\sqrt{41}[/tex] is irrational.
Hence, The given function has 1 rational root and 2 irrational roots.
