The function is

.
To the left of 1 the function is a quadratic polynomial, to the right, it is a linear polynomial. Polynomial functions are always continuous, so the only candidate point for discontinuity is x=1.
The left limit is calculated with 1 substituted in

, which gives 5.
The right limit, is computed using the rule for the right part of 1, that is x+4.
Thus, the right limit is 1+4=5.
So, both left and right limits are equal. Now if f(1) is 5, then the function is continuous at 1.
But the function is not defined for x=1, that is x=1 is not in the domain of the function. Thus, we have a "whole" (a discontinuity) in the graph of the function.
The reason is now clear:
Answer:<span> f(1) is not defined</span>
Answer:
If:
(1)
And we have to rewrite the following expresion in terms of
:
(2)
Firstly we have to rearrange the right side of equation (2) in terms of
, factorizing by the common factor:
Then, we can substitute
by
:
And set it equal to zero:
>>>>This is the resulting equation
Now, each term of an algebraic expresion (like the equation above) is composed of sign, coefficient, variable and exponent. The terms are separated from each other by the plus sign (+) or the minus sign (-).
In this case, the variable is
and the number that multiplies the variable (the coefficient) is
, whereas the constant (which is the term with no variable) is 