The other zero of the given function is [tex](\frac{11}{3}, 0)[/tex] .
How to find the x-intercept of a quadratic function ?
Let us assume f(x) [tex]=ax^{2} +bx+c[/tex], then to find the x-intercepts of this quadratic equation, let y = 0, i.e. [tex]ax^{2} +bx+c=0[/tex]
How to find zeros of the given function ?
Given, f(x)= [tex]3x^{2} -5x-22[/tex], whose one zero is (-2,0)
Now, to find the x-intercepts of the given quadratic equation, we have to take y = 0.
∴ [tex]3x^{2} -5x-22=0[/tex]
Now we have to solve this equation to find it's another zero.
[tex]3x^{2} -5x-22=0[/tex]
⇒ [tex]3x^{2} -(11-6)x-22=0[/tex]
⇒ [tex]3x^{2} -11x+6x-22=0[/tex]
⇒ [tex]x(3x-11)+2(3x-11)=0[/tex]
⇒ [tex](3x-11)(x+2)=0[/tex]
We know that, if multiplication of two terms is zero, then each of this term is seperately zero.
∴ [tex]3x-11=0[/tex] or [tex]x+2=0[/tex]
⇒ [tex]3x=11[/tex] or [tex]x=-2[/tex]
⇒ [tex]x=\frac{11}{3}[/tex] or [tex]x=-2[/tex]
So, another zero of the function is [tex](\frac{11}{3}, 0)[/tex].
Learn more about zeros of a quadratic equation here :
https://brainly.com/question/1498156
#SPJ2
