The equation provided in the image is $y = x^2 - x + 1$, and the task asks to rewrite the quadratic equation in vertex form and to find the vertex.

Step-by-Step Solution:

1. Standard Form of a Quadratic Function: The equation is in the standard form $y = ax^2 + bx + c$, where: $a = 1, \quad b = -1, \quad c = 1$ To convert this to vertex form $y = a(x - h)^2 + k$, we need to complete the square.

2. Complete the Square: Start by isolating the $x$-terms: $y = x^2 - x + 1$ To complete the square, take half of the coefficient of $x$, which is $-1/2$, and square it: $\left(\frac{-1}{2}\right)^2 = \frac{1}{4}$ Now, add and subtract $\frac{1}{4}$ inside the equation: $y = \left(x^2 - x + \frac{1}{4}\right) + 1 - \frac{1}{4}$ This simplifies to: $y = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}$

3. Vertex Form: Now the equation is in vertex form: $y = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}$ The vertex form is $y = a(x - h)^2 + k$, where $h = \frac{1}{2}$ and $k = \frac{3}{4}$.

4. Vertex: The vertex of the parabola is $\left(\frac{1}{2}, \frac{3}{4}\right)$.

Final Answer:

• Vertex form: $y = \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}$
• Vertex: $\left(\frac{1}{2}, \frac{3}{4}\right)$

Would you like a further explanation or have any other questions?

Related Questions:

1. How can we determine the vertex from the standard form without completing the square?
2. What is the significance of the vertex in a quadratic graph?
3. How do we determine whether the parabola opens upwards or downwards?
4. Can we find the axis of symmetry directly from the equation?
5. How do we find the y-intercept of a quadratic function?

Tip:

To quickly find the vertex of a parabola in standard form $ax^2 + bx + c$, use the formula $h = \frac{-b}{2a}$ for the x-coordinate of the vertex.