Domain of y= (x²+3x+1)÷(x²-16)

Domain is the set of all possible values at which function is defined. In this problem, function is undefined when denominator is zero, here we can see denominator is zero at x=+4 and x=-4, so domain is set of all real numbers except 4 and -4. Regards Dr Mohammad Suleman Quraishi

X= -b+√b2 -4ac /2a X = -3+ √9 - 4×1×1/2×1 X = -3 + √5/2 Y = -3+√5/2/ (X-4) (x+4)

The domain of a function is the set of all possible input values (x-values) for which the function is defined and produces a valid output value (y-value). In other words, it's the set of all x-values for which the function is defined and does not produce an undefined output, such as a division by zero. For the function y = (x²+3x+1)÷(x²-16), the denominator (x²-16) is equal to zero when x² = 16, therefore x = ±4. Therefore, the domain of the function is all real numbers except x = 4 and x = -4. This can be written as: {x|x ∈ R, x≠4, x≠-4} In summary, the domain of the function y = (x²+3x+1)÷(x²-16) is all real numbers except x = 4 and x = -4 and can be represented by the set {x|x ∈ R, x≠4, x≠-4}

Domain of this function means for which values of x this function y would be defined or have real values. Any number cannot be divided by 0 so x^2-16 cannot be 0. So x cannot be the square root of 16 so it cant be either 4 or -4. The domain is all real values of x except for 4 or -4.

to find the domain of this function, you only need to look into the denominator of the function ie. (x^2-16) is not zero So if the denominator is zero then, x^2-16=0 x^2=16 therefore, x=+4 or x=-4. so the domain can be any real number except +4 and -4.

The domain of a function is the set of all possible values of the input variable (x) for which the function is defined. In this case, the function is: y = (x² + 3x + 1) / (x² - 16) To find the domain, we need to identify any values of x that would make the denominator (x² - 16) equal to zero, since division by zero is undefined. Setting the denominator to zero, we get: x² - 16 = 0 Solving for x, we get: x = ±4 Therefore, the function is undefined for x = 4 and x = -4, since they would make the denominator zero. Therefore, the domain of the function is all real numbers except for x = 4 and x = -4. In interval notation, we can write the domain as: (-∞, -4) ∪ (-4, 4) ∪ (4, ∞)

Using almighty formula, ax2+ bx+c, X =( -b + or - √b^2 - 4ac)/2a and using difference of two squares a^2 - b^2 = {a-b)(a+b) X^2-16= x^2-4^2= {x+4)(x-4) X^2+3x+1 = ax^2+bx+c, will make a= 1, b=3 and c=1 X = (-b+or-√b^2-4ac)/2a X= (-3 + √3^2-4*1*1)/2x1 OR (-3-√3^2-4*1*1)/2x1 = (-3+√9-4)/2 OR(-3-√9-4)/2 =(-3+√5)/2 OR(-3-√5)/2 FINAL ANSWER= =(-3+√5)/2(x+4)(x-4) OR(-3-√5)/2(x-4)(x+4)