Please help me solve and find the excluded values for this problem. Also please show me the work for this problem I want to learn! h(x)=x^2-9/x^2-4x+3
same idea as before, factor first, then divide, restrict what you divided by
Again, I am putting in the brackets, they are extremely important here
h(x)=(x^2-9)/(x^2-4x+3)
= (x+3)(x-3)/(x-1)(x-3))
= (x+3)/(x-1) , x ≠ 3
To find the excluded values for the function h(x) = (x^2-9)/(x^2-4x+3), we need to identify the values of x that make the denominator equal to zero. Division by zero is undefined in mathematics, so these values should be excluded from the domain of the function.
To start, set the denominator (x^2-4x+3) equal to zero:
x^2 - 4x + 3 = 0
Next, we can factor the quadratic equation:
(x-3)(x-1) = 0
Now, set each factor equal to zero and solve for x:
x-3 = 0 --> x = 3
x-1 = 0 --> x = 1
So the excluded values for the function h(x) occur at x = 3 and x = 1, as they make the denominator equal to zero.
To check our answer, we can also sketch the graph of the function:
First, find the vertical asymptotes: The excluded values will be where the graph has vertical asymptotes. In this case, the vertical asymptotes will be x = 3 and x = 1.
Next, find the horizontal asymptote: By analyzing the degree of the numerator and denominator, we can see that their degrees are equal (both 2). So, the horizontal asymptote will be found by looking at the ratio of the leading coefficients. The leading coefficient of x^2-9 is 1, and the leading coefficient of x^2-4x+3 is also 1. Thus, the horizontal asymptote is y = 1.
Finally, plot a few points and sketch the graph, ensuring the function approaches the vertical asymptotes and the horizontal asymptote as x approaches positive and negative infinity, respectively.
Considering the explanation and provided work, we have identified the excluded values of x = 3 and x = 1 for the function h(x) = (x^2-9)/(x^2-4x+3).