How do I solve #f(x)=(x^3-8)/(x^2+5x+6)# on a TI-84?

1 Answer
Apr 27, 2018

zero #= (2, 0)#

Explanation:

Given: #f(x) = (x^3 - 8)/(x^2 + 5x + 6)#

On a TI-84 select the #Y=# key and enter in the function

#" " #Y1 = (x^3 - 8)/(x^2 + 5x + 6)

Make sure your WINDOW is in 6:ZStandard

2nd QUIT (above MODE)

GRAPH key (lower right of screen)

To find the solution you are finding #x#-intercepts (zeros).

2ND CALC (TRACE key) 2:zero

Left Bound? Select a point below the x-axis using the arrow keys, then hit ENTER

Right Bound? Select a point above the x-axis using the arrow keys, then hit ENTER

Guess? ENTER (you can move closer to the #x#-intercept, but it isn't necessary unless there are two zeros close to each other).

Zero #= (2, 0)#