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)
Sure! Let's work through each problem step by step:
1) To find the values that would be left out of the range of f, we need to determine if there are any values of x that would result in the denominator of the function being equal to zero. This is because division by zero is undefined. So our first step is to set the denominator equal to zero and solve for x:
4x^2 + 12x - 72 = 0
Now we can factor this quadratic equation:
4(x^2 + 3x - 18) = 0
(x + 6)(x - 3) = 0
So we have two possible values for x: x = -6 and x = 3. These are the values that would make the denominator zero, and therefore they would be left out of the range of f.
Answer: -6, 3
2) To find the y-intercept of f, we need to find the value of y when x = 0. Substituting x = 0 into the equation:
f(0) = (6*0 - 3)/(2*0 + 1)
= -3/1
= -3
So the y-intercept is the point (0, -3).
Answer: (0, -3)
3) To find the equation for any vertical asymptotes of f, we need to determine if there are any values of x that would make the denominator of the function equal to zero, but not the numerator. This is because division by zero is undefined. So our first step is to set the denominator equal to zero and solve for x:
2x^2 - 3x = 0
Now we can factor this quadratic equation:
x(2x - 3) = 0
So we have two possible values for x: x = 0 and x = 3/2. These values would make the denominator zero, but not the numerator, creating vertical asymptotes at x = 0 and x = 3/2.
Answer: x = 0, x = 3/2
4) To find the x-intercept of f, we need to find the value(s) of x when y = 0. Setting the numerator equal to zero:
4/(x - 6) = 0
Since a fraction is zero only when the numerator is zero and the denominator is non-zero, we know that the numerator must equal zero:
4 = 0 (No solution)
Since the equation has no solution, there are no x-intercepts for this function.
Answer: none
5) To find any holes in the graph of f, we need to set the numerator and denominator equal to zero and solve for x:
x^2 + x - 12 = 0
(x + 4)(x - 3) = 0
x^2 + 3x + 2 = 0
(x + 1)(x + 2) = 0
From these equations, we find that x = -4 and x = 3 would create holes in the graph at those respective x-values. To find the y-values for these holes, we substitute these x-values back into the original function:
When x = -4:
f(-4) = (-4^2 + (-4) - 12)/(-4^2 + 3(-4) + 2)
= (16 - 4 - 12)/(16 - 12 + 2)
= 0
When x = 3:
f(3) = (3^2 + 3 - 12)/(3^2 + 3(3) + 2)
= (9 + 3 - 12)/(9 + 9 + 2)
= 0
Therefore, the graph has holes at the points (-4, 0) and (3, 0).
Answer: (-4, 0), (3, 0)
6) To find any values left out of the domain of f, we need to determine if there are any values of x that would make the denominator of the function equal to zero. This is because division by zero is undefined. So our first step is to set the denominator equal to zero and solve for x:
x^2 - 8x + 15 = 0
Now we can factor this quadratic equation:
(x - 5)(x - 3) = 0
So we have two possible values for x: x = 5 and x = 3. These values would make the denominator zero, and therefore they would be left out of the domain of f.
Answer: 5, 3