Question

Marsha wants to determine the vertex of the quadratics vertex of the quadratics functions f(x)=x^2-x+2

Answer 1

Answer:

The vertex of the function is $(\frac{1}{2}, \frac{7}{4})$

Step-by-step explanation:

Suppose we have a quadratic equation in the following format:

$y = f(x) = ax^{2} + bx + c$

The vertex of the function is the point $(x_{v}, f(x_{v})$, in which

$x_{v} = -\frac{b}{2a}$

In this problem:

$f(x) = x^{2} - x + 2$

This means that $a = 1, b = -1, c = 2$

So

$x_{v} = -\frac{-1}{2} = \frac{1}{2}$

$f(x_{v}) = f(\frac{1}{2}) = (\frac{1}{2})^{2} - \frac{1}{2} + 2 = \frac{1}{4} - \frac{1}{2} + 2 = \frac{1 - 2 + 8}{4} = \frac{7}{4}$

The vertex of the function is $(\frac{1}{2}, \frac{7}{4})$