Sketch the graph of the following polynomial. (Label all x intercepts, y intercepts, maxima, minima, and identify the range and domain.)
f(x)= 1/2( x^2-2x+15)
Check here:
http://www.wolframalpha.com/input/?i=1%2F2+(+x%5E2-2x%2B15)
Use what you know about quadratics to figure out what it shows. You can play around with other functions and see how things work...
To sketch the graph of the given polynomial function, f(x) = 1/2(x^2 - 2x + 15), we can follow a few steps:
Step 1: Find the x-intercepts
To determine the x-intercepts, we set f(x) equal to zero and solve for x. In this case, we have:
0 = 1/2(x^2 - 2x + 15)
Now, we can solve the quadratic equation x^2 - 2x + 15 = 0 using factoring, completing the square, or the quadratic formula. In this case, factoring doesn't work, so we'll use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
where a = 1/2, b = -2, and c = 15.
Substituting these values, we have:
x = (-(-2) ± √((-2)^2 - 4(1/2)(15))) / 2(1/2)
x = (2 ± √(4 - 30)) / 1
x = (2 ± √(-26)) / 1
Since √(-26) is imaginary (there is no real square root of negative numbers), the quadratic equation has no real x-intercepts. Therefore, there are no x-intercepts for the graph.
Step 2: Find the y-intercept
To determine the y-intercept, we set x equal to zero and evaluate f(x).
f(0) = 1/2(0^2 - 2(0) + 15)
f(0) = 1/2(0 - 0 + 15)
f(0) = 1/2(15)
f(0) = 15/2
So, the y-intercept is (0, 15/2).
Step 3: Determine the vertex (maximum or minimum)
To find the vertex, we can use the formula x = -b/2a. For our quadratic equation x^2 - 2x + 15 = 0, we have a = 1/2 and b = -2.
x = -(-2) / 2(1/2)
x = 2 / 1
x = 2
To find the corresponding y-value (f(x)) for the vertex, substitute x = 2 into the original equation:
f(2) = 1/2(2^2 - 2(2) + 15)
f(2) = 1/2(4 - 4 + 15)
f(2) = 1/2(15)
f(2) = 15/2
So, the vertex is at (2, 15/2).
Step 4: Determine the range and domain
Since the coefficient of x^2 is positive (1/2), the graph opens upwards, indicating a minimum. Therefore, the graph has a minimum value at the vertex, which is 15/2.
The range (y-values) of the graph is all real numbers greater than or equal to the minimum value of 15/2. In interval notation, the range is [15/2, ∞).
The domain (x-values) is all real numbers since there are no restrictions or excluded values.
Finally, we can sketch the graph of the polynomial f(x) = 1/2(x^2 - 2x + 15), which is an upward-opening parabola with no x-intercepts, a y-intercept at (0, 15/2), and a minimum point at the vertex (2, 15/2).