answer the following for the given quadratic function
f(x)=x^2-2x+1
a. Does the graph of f open up or down?
b. What is the vertex of (h,k) of f?
c. What is the axis of symmetry?
D. What is the intercepts?
E. What is the domain?
F. What is the range?
G. On what interval is f increasing?
H. On what interval is f decreasing?
please show work
y=(x-1)^2
now just run down the list of stuff you know, This is mostly algebra I.
To answer these questions about the quadratic function f(x) = x^2 - 2x + 1, we can apply various concepts related to quadratic functions. Let's go through each question and explain how to find the answers:
a. To determine whether the graph of f opens up or down, we need to look at the coefficient of the x^2 term, which is 1 in this case. Since this coefficient is positive (+1), the graph opens upward.
b. The vertex of a quadratic function can be found using the formula: h = -b/(2a), where a and b are the coefficients of the quadratic terms. In our case, a = 1 and b = -2. Plugging these values into the formula, we get: h = -(-2)/(2*1) = 2/2 = 1.
To find the y-coordinate (k) of the vertex, we substitute the value of h into the function: k = f(1) = (1)^2 - 2(1) + 1 = 1 - 2 + 1 = 0. Therefore, the vertex of f is (1, 0).
c. The axis of symmetry for a quadratic function is a vertical line that passes through the vertex. In this case, since the x-coordinate of the vertex is 1, the equation of the axis of symmetry is x = 1.
d. To find the x-intercepts, we set f(x) = 0 and solve the quadratic equation. So, for f(x) = x^2 - 2x + 1, we set x^2 - 2x + 1 = 0.
Using factoring, we can rewrite the equation as (x - 1)(x - 1) = 0. Thus, the equation is satisfied when x = 1. Therefore, the quadratic function intersects the x-axis at the point (1, 0), which we already identified as the vertex.
e. The domain of a quadratic function is the set of all real numbers for which the function is defined. In this case, since no values of x are restricted, the domain of f(x) = x^2 - 2x + 1 is (-∞, +∞), which means all real numbers.
f. The range of a quadratic function is the set of all possible y-values the function can have. In this case, since the graph opens upward, the minimum y-value occurs at the vertex, which we found to be (1, 0). Thus, the range of f(x) = x^2 - 2x + 1 is [0, +∞), which means all y-values greater than or equal to 0.
g. To determine where the function is increasing, we need to find the intervals where the function's slope (or derivative) is positive. The slope of a quadratic function is represented by the coefficient of the linear term, which is -2 in this case. Since the slope is negative (-2), the function is decreasing for all x values.
h. Similarly, to determine where the function is decreasing, we need to find the intervals where the function's slope (or derivative) is negative. As mentioned above, since the slope is always negative (-2), the function is decreasing for all x values.
In summary:
a. The graph of f opens upward.
b. The vertex of f is (1, 0).
c. The axis of symmetry is x = 1.
d. The function has one x-intercept at (1, 0).
e. The domain is (-∞, +∞).
f. The range is [0, +∞).
g. The function is not increasing over any interval.
h. The function is decreasing over all x-values.