Respuesta :
The slopes of the secant lines do not appear to be approaching a limit.
For each value of x, calculate y and then use the slope formula to find the slope of P Q for the given limit.
x = 2 → y = 0 : Q(2, 0) ⇒ m =0 − 0 / 2 − 1 = 0
x = 1.5 → y ≈ 0.866 : Q(1.5, 0.866) ⇒ m = 0.866 − 0/ 1.5 − 1 ≈ 1.73
x = 1.4 → y ≈ −0.434 : Q(1.4, −0.434) ⇒ m =−0.434 − 0 / 1.4 − 1 = −1.08
x = 1.3 → y ≈ −0.823 : Q(1.3, −0.823) ⇒ m = −0.823 − 0/1.3 − 1 = −2.74
x = 1.2 → y ≈ 0.866 : Q(1.2, 0.866) ⇒ m = 0.866 − 0 / 1.2 − 1 = 4.33
x = 1.1 → y ≈ −0.282 : Q(1.1, −0.282) ⇒ m = −0.282 − 0/1.1 − 1 = −2.82
x = 0.9 → y ≈ −0.342 : Q(0.9, −0.342) ⇒ m = −0.342 − 0/0.9 − 1 = 3.42
x = 0.8 → y = 1 : Q(0.8, 1) ⇒ m = 1 − 0/ 0.8 − 1 = −5
x = 0.7 → y ≈ 0.782 : Q(0.7, 0.782) ⇒ m = 0.782 − 0/0.7 − 1 = −2.61
x = 0.6 → y ≈ 0.866 : Q(0.6, 0.866) ⇒ m = 0.866 − 0 / 0.6 − 1 = −2.17
x = 0.5 → y = 0 : Q(0.5, 0) ⇒ m = 0 − 0 / 0.5 − 1 = 0
Hence, The slopes of the secant lines do not appear to be approaching a limit.
To know more about slopes check the below link:
https://brainly.com/question/3493733
#SPJ4
