Question

Marsha wants to determine the vertex of the quadratic function f(x) = x2 – x + 2. What is the function’s vertex?

(one-half, seven-quarters)
(one-half, three-halves)
(1, 1)
(1, 3)

Answer 1

To find the vertex of the quadratic function \( f(x) = x^2 - x + 2 \), we can use the vertex formula for a quadratic function in the form \( f(x) = ax^2 + bx + c \). The x-coordinate of the vertex is given by:

\[
x = -\frac{b}{2a}
\]

Here, \( a = 1 \) and \( b = -1 \).

Substituting the values into the formula:

\[
x = -\frac{-1}{2 \cdot 1} = \frac{1}{2}
\]

Now, we need to find the y-coordinate of the vertex by substituting \( x = \frac{1}{2} \) back into the function \( f(x) \):

\[
f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) + 2
\]

Calculating this step-by-step:

1. \( \left(\frac{1}{2}\right)^2 = \frac{1}{4} \)
2. \( -\left(\frac{1}{2}\right) = -\frac{1}{2} = -\frac{2}{4} \) (convert to have a common denominator)
3. Adding these:
\[
f\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{2}{4} + 2
\]
4. \( 2 = \frac{8}{4} \) (convert to have a common denominator)
5. Now calculating:
\[
f\left(\frac{1}{2}\right) = \frac{1}{4} - \frac{2}{4} + \frac{8}{4} = \frac{1 - 2 + 8}{4} = \frac{7}{4}
\]

Thus, the vertex of the quadratic function \( f(x) = x^2 - x + 2 \) is

\[
\left(\frac{1}{2}, \frac{7}{4}\right)
\]

Therefore, the correct answer is:

**(one-half, seven-quarters)**