Question

H(x)= ​(x−5) ​2 ​​ +4(x−5)+4 ​ ​1 ​​ space, h, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, left parenthesis, x, minus, 5, right parenthesis, start superscript, 2, end superscript, plus, 4, left parenthesis, x, minus, 5, right parenthesis, plus, 4, end fraction for what value of xxx is the function hhh above undefined?

Answer 1

The function is undefined when x = 3. You have a function where the numerator is 1 and the denominator is an expression with the variable x. That function will be undefined at those points where the denominator has a value of zero. So let's look at just that part of the function and solve it for where its value is zero. (x-5)^2 + 4(x-5) + 4 = 0 Perform the squaring of (x-5) x^2 - 10x + 25 + 4(x-5) + 4 = 0 Distribute the 4 x^2 - 10x + 25 + 4x - 20 + 4 = 0 Combine common terms (add -10x to 4x, and add 25 to -20 to 4) x^2 - 6x + 9 = 0 We now have a quadratic equation with a=1, b = -6, and c=9. Solve using the quadratic formula, giving a root of x = 3. Verify by seeing if x=3 gives a denominator of 0.. (x-5)^2 + 4(x-5) + 4 = y (3-5)^2 + 4(3-5) + 4 = y (-2)^2 + 4(-2) + 4 = y 4 + -8 + 4 = y 0 = y

Answer 2

At the value of x = 3 the function H(x) is undefined.

Given :

$H(x) = \dfrac{1}{(x-5)^2+4(x-5)+4}$

Solution :

We know that for the values of x where denominator is zero the function H(x) is not define.

$(x-5)^2+4(x-5)+4\neq 0$

$x^2-6x+9 \neq 0$

$x^2-3x-3x+9\neq 0$

$(x-3)^2\neq 0$

$x\neq 3$

At x = 3 the function H(x) is undefined.

For more information, refer the link given below

https://brainly.com/question/13911928