Question

What is the discontinuity and zero of the function f(x) = the quantity of 3 x squared plus x minus 4, all over x minus 1?

Answer 1

Discontinuity at (-1, 1), zero at (four thirds, 0)

Discontinuity at (-1, 1), zero at (negative four thirds , 0)

Discontinuity at (1, 7), zero at (four thirds , 0)

Discontinuity at (1, 7), zero at (negtive four thirds, 0)

Answer 2

\(\Large f(x) = \dfrac{3x^2+x-4}{x-1}\)

Answer 3

to get the zeros, put f(x) = 0

Answer 4

so, numerator = 0,
\(3x^2+x-4 = 0\)
xan you solve this quadratic ?

Answer 5

*can

Answer 6

let me see tell me if its right

Answer 7

I have so far 9 + x - 4=0

Answer 8

9 ?
you've solve quadratics before ?
using factor method

Answer 9

whats next

Answer 10

Don't I do 3 times 3?

Answer 11

but thats not correct...

\(3x^2 +x-4 = 0 \\ 3x^2 -3x +4x-4 = 0 \\ 3x (x-1)+4(x-1)=0\)

Answer 12

ever done such thing ^^ ?

Answer 13

ohh no lol

Answer 14

\(3x^2 +x-4 = 0 \\ 3x^2 -3x +4x-4 = 0 \\ 3x (x-1)+4(x-1)=0 \\ (3x+4)(x-1) = 0\)

Answer 15

thats called factoring the quadratic!

Answer 16

see if you get the steps, ask if any doubts..

Answer 17

ok I will

Answer 18

whered the one come from in 3x(x - 1)

Answer 19

\(3x^2 = 3\times x \times x \\ -3x = -3 \times x\)
so 3 and 'x' are common, factoring them out!
\(3\times x (x -1) = 3x (x-1)\)

Answer 20

got it ?

Answer 21

going forward,

\((3x+4)(x-1) =0 \\ 3x+4 = 0 \quad or \quad x-1= 0\)
but x-1 cannot be = 0 , since denominator cannot be = 0

so only
3x+4 = 0
what do you get x from here ?

Answer 22

the function will be discontinuous at points where denominator = 0
so, for discontinuity
solve x-1= 0, x =... ?