Question

X^(2)+4x+5 in the form a(x - h)^(2)+k

Answer 1

1×(x - -2)^2 + 1

Explanation:

x^2 had no multiplying number (or just simple "1").

therefore, a has to be 1.

there are only positive terms, so h has to be negative (so that - - becomes + in the addition).

5 is not the square of an integer number, so k has to be >0

the only number between 1 and 5 that is a square of an integer number is 4.

so, the assumption is that h = -2 (remember, negative and the square root of 4). then k has to be 1.

check :

1×(x - -2)^2 + 1 = (x + 2)^2 +1 = x^2 + 2×2×x + 2^2 + 1 =

= x^2 + 4×x + 4 + 1 = x^2 + 4×x + 5

correct.