# f(t)=t^2+5t-7 a. find t so that f(t)=-1 b.find f(k+3)

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### 2 Answers

`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` .*

**Sources:**

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