f(t)=t^2+5t-7 a. find t so that f(t)=-1 b.find f(k+3)
2 Answers
jeew-m | College Teacher | (Level 1) Educator Emeritus
`f(t) = t^2+5t-7`
a)
`f(t) = -1`
`t^2+5t-7 = -1`
`t^2+5t-6 = 0`
`t^2+6t-t-6 = 0`
`t(t+6)-1(t+6) = 0`
`(t+6)(t-1) = 0`
`t = 1` or `t = -6`
So the answers are t = 1 and t = -6.
b)
to find f(k+3) we need to replace t with (k+3).
`f(t)=t^2+5t-7`
`f(k+3) = (k+3)^2+5(k+3)-7`
`f(k+3) = k^2+11k+17`
So the answer is `f(k+3) = k^2+11k+17` .
tonys538 | Student, Undergraduate | (Level 1) Valedictorian
The function f(t)=t^2+5t-7.
To determine the value of t for which f(t) = -1, solve the equation
t^2 + 5t - 7 = -1
t^2 + 5t - 6 = 0
t^2 + 6t - t - 6 = 0
t(t + 6) - 1(t +6) = 0
(t - 1)(t + 6) = 0
t = 1 and t = -6
For t = 1 and t = -6, the value of f(t) is -1.
f(k+3)
= (k+3)^2+5(k+3)-7
= k^2 + 6k + 9 + 5k + 15 - 7
= k^2 + 11k + 17