find the domain, range, y and x-intercept, vertical asymptote, maximum, minimum, one-to-one, and horizontal asymptote of:
a. f(x)=x^2-6x+5
b. f(x)=x^2-6x+5 on domain [-1,6]
all polynomials have domain of all reals
No polynomial has any kind of asymptote
x^2-6x+5 = (x-3)^2 - 4
So you can find the vertex, which is a minimum.
Since the vertex is farther away from x=-1 than it is from x=6, f(-1) > f(6)
Since f(3-x) = f(3+x), it is evidently not 1:1.
To find the domain, range, y-intercept, x-intercept, vertical asymptote, maximum, minimum, one-to-one, and horizontal asymptote of a function, we will go step by step for each function mentioned.
a. f(x) = x^2 - 6x + 5
1. Domain: The domain of a function is the set of all possible x-values for which the function is defined. For polynomial functions like this, the domain is all real numbers since there are no restrictions or divisions by zero. So, the domain is (-∞, ∞).
2. Range: The range is the set of all possible y-values that the function can take. For this quadratic function, the range depends on the vertex of the parabola. Since the coefficient of the x^2 term is positive (+1), the parabola opens upwards, and the vertex represents the minimum point. Therefore, the range is [y-coordinate of the vertex, +∞).
To find the y-intercept, we substitute x = 0 into the equation:
f(0) = 0^2 - 6(0) + 5
f(0) = 0 - 0 + 5
f(0) = 5
So, the y-intercept is (0, 5).
To find the x-intercept(s), we set f(x) = 0:
0 = x^2 - 6x + 5
By factoring or using the quadratic formula, we find that the x-intercepts are (1, 0) and (5, 0).
To find the vertical asymptote(s), we need to check if there are any x-values for which the function approaches infinity or negative infinity. However, for this quadratic function, since the degree of the polynomial is 2, there are no vertical asymptotes.
To find the maximum or minimum points, we can notice that the quadratic function is in the form f(x) = ax^2 + bx + c, where a > 0. In this case, a = 1, and since a > 0, there is a minimum point. The x-coordinate of the vertex can be found by the formula x = -b / (2a), which is the axis of symmetry.
x = -(-6) / (2 * 1)
x = 6/2
x = 3
To find the y-coordinate of the vertex, substitute the value of x = 3 into the equation:
f(3) = 3^2 - 6(3) + 5
f(3) = 9 - 18 + 5
f(3) = -4
So, the vertex is located at (3, -4).
Since this function is a quadratic function and has a vertex, it is not a one-to-one function.
Lastly, for quadratic functions, there is no horizontal asymptote since the function does not approach a particular y-value as x tends to infinity or negative infinity.
b. f(x) = x^2 - 6x + 5 on the domain [-1,6]
For part b, the domain is specified as [-1, 6], which means we consider only the x-values between -1 and 6 (inclusive).
The process for finding the range, y-intercept, x-intercepts, vertical asymptote, maximum, minimum, one-to-one, and horizontal asymptote remains the same as mentioned above for function a. However, the range, x-intercepts, and maximum/minimum points might change within the specified domain.
To find the range, y-intercept, x-intercepts, vertex, and other details within the specified domain [-1, 6], you can follow the same steps as mentioned for function a, but make sure to restrict your calculations to the specified domain.