I need help understanding how to do these problems. Thanks!
1) Find the value(s) that would be left out of the range of f. If there is more than one value separate them by commas. Leave fractional answers in fraction form. If there are no values left out of the range respond by typing "none".
f(x) = (x^2+4x-12)/(4x^2+12x-72)
2)Find the y-intercept of f. Write the y-intercept as an ordered pair in simplified form. If there is more than one y-intercept separate each ordered pair by a comma. Write all fractional answers as fractions. If there is not a y-intercept respond by typing "none".
f(x) = (6x-3)/(2x+1)
3)Find the equation for any vertical asymptote(s) of f. Leave any fraction answers in fractional form.
f(x) = (8x^2+3x)/(2x^2-3x)
4) Find the x-intercept of f. Write the x-intercept as an ordered pair. If there is more than one x-intercept separate each ordered pair by a comma. Write all fractional answers as fractions. If there is not an x-intercept respond by typing "none".
f(x) = 4/(x-6)
5) Find any holes of the graph of f. Write the hole(s) as an ordered pair in simplified form. Leave fraction answers in fraction form. If there is more than one hole separate each ordered pair by a comma. If there is not any holes in the graph of f respond by typing "none".
f(x) = (x^2+x-12)/(x^2+3x+2)
6)List any values that are left out of the domain for f. Leave any fractional answers in fraction form. If there is more than one value separate them by commas.
f(x) = (x^2-4x-5)/(x^2-8x+15)
1) To find the values that would be left out of the range of f, we need to determine the values that would make the denominator of the fraction equal to zero. This is because dividing by zero is undefined.
The denominator of f(x) is 4x^2 + 12x - 72. We can find the values that would make this equal to zero by solving the equation:
4x^2 + 12x - 72 = 0
We can factor the equation to:
4(x^2 + 3x - 18) = 0
Now we can solve for x by setting each factor equal to zero:
x^2 + 3x - 18 = 0
Using factoring, we can rewrite the equation as:
(x + 6)(x - 3) = 0
Setting each factor equal to zero gives us the possible values for x:
x + 6 = 0 or x - 3 = 0
Solving these equations gives us:
x = -6 or x = 3
Therefore, the values -6 and 3 would be left out of the range of f.
Answer: -6, 3
2) To find the y-intercept of f, we need to determine the value of f(x) when x is equal to zero. This gives us the y-coordinate of the point where the graph of f intersects the y-axis.
In the function f(x) = (6x - 3)/(2x + 1), we can substitute x = 0 and solve for f(0):
f(0) = (6(0) - 3)/(2(0) + 1)
= (-3)/(0 + 1)
= -3/1
= -3
Therefore, the y-intercept of f is the point (0, -3).
Answer: (0, -3)
3) To find the equation for any vertical asymptotes of f, we need to determine the values of x that would make the denominator equal to zero. This is because dividing by zero results in an undefined value, causing the graph of f to approach infinity.
In the function f(x) = (8x^2 + 3x)/(2x^2 - 3x), we set the denominator equal to zero:
2x^2 - 3x = 0
We can factor out x to get:
x(2x - 3) = 0
Setting each factor equal to zero gives us the possible values for x:
x = 0 or 2x - 3 = 0
Solving the second equation gives us:
2x = 3
x = 3/2
Therefore, the vertical asymptote of f is x = 0 and x = 3/2.
Answer: x = 0, 3/2
4) To find the x-intercept of f, we need to determine the values of x that would make the numerator equal to zero. This is because when the numerator is zero, the function value is zero and the graph intersects the x-axis.
In the function f(x) = 4/(x - 6), we set the numerator equal to zero:
4 = 0
Since there is no value of x that would make 4 equal to zero, there are no x-intercepts for f.
Answer: none
5) To find any holes of the graph of f, we need to determine the values of x that would make both the numerator and the denominator equal to zero.
In the function f(x) = (x^2 + x - 12)/(x^2 + 3x + 2), we set both the numerator and the denominator equal to zero:
x^2 + x - 12 = 0
(x - 3)(x + 4) = 0
x - 3 = 0 or x + 4 = 0
x = 3 or x = -4
Both x = 3 and x = -4 would make the numerator and the denominator equal to zero. This means that there are holes in the graph of f at the points (3, undefined) and (-4, undefined).
Answer: (3, undefined), (-4, undefined)
6) To find the values that are left out of the domain of f, we need to determine the values of x that would make the denominator equal to zero. Dividing by zero is undefined, so we need to exclude those values from the domain of f.
In the function f(x) = (x^2 - 4x - 5)/(x^2 - 8x + 15), we set the denominator equal to zero:
x^2 - 8x + 15 = 0
We can factor the equation to:
(x - 3)(x - 5) = 0
Setting each factor equal to zero gives us the possible values for x:
x - 3 = 0 or x - 5 = 0
Solving these equations gives us:
x = 3 or x = 5
Therefore, the values 3 and 5 are left out of the domain of f.
Answer: 3, 5