Question

How do you find the zeros of the function `f(x)=(20x^2+11x-40)/(2x+5)`?

Answer 1

`x~~-1.72,x~~1.17" to 2 dec. places"`

Explanation:

`"the denominator of f(x) cannot be zero as this would "`
`"make f(x) undefined"`

`"the zeros are found by equating the numerator to zero"`

`rArr"solve "20x^2+11x-40=0larrcolor(blue)"standard form"`

`"solve using the "color(blue)"quadratic formula"`

`"with "a=20,b=11" and "c=-40`

`•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)`

`rArrx=(-11+-sqrt(121+3200))/40`

`color(white)(rArrx)=(-11+-sqrt3321)/40=(-11+-9sqrt41)/40`

`rArrx=-11/40+-(9sqrt41)/40`

`rArrx=-11/40-(9sqrt41)/40" or "x=-11/40+(9sqrt41)/40`

`rArr"zeros are "x~~-1.72,x~~1.17" to 2 dec. places"`