How do you find the vertex and intercepts for y = x^2 - x + 2 ?

Jan 12, 2016

If $y = f \left(x\right) = {x}^{2} - x + 2$, the $y$-intercept is $f \left(0\right) = 2$, the vertex can be found by completing the square (shown below) to be at the point $\left(\frac{1}{2} , f \left(\frac{1}{2}\right)\right) = \left(\frac{1}{2} , \frac{7}{4}\right)$, and the quadratic formula can be used to say there are no $x$-intercepts.

Explanation:

Here's the method of completing the square to find the vertex form of $f$:

$f \left(x\right) = {x}^{2} - x + 2 = {x}^{2} - x + {\left(- \frac{1}{2}\right)}^{2} + 2 - {\left(- \frac{1}{2}\right)}^{2}$

$= {\left(x - \frac{1}{2}\right)}^{2} + 2 - \frac{1}{4} = {\left(x - \frac{1}{2}\right)}^{2} + \frac{7}{4}$.

This shows that the vertex, or low point of this upward pointing parabola, is at $\left(x , y\right) = \left(\frac{1}{2} , f \left(\frac{1}{2}\right)\right) = \left(\frac{1}{2} , \frac{7}{4}\right)$.

For the $x$-intercepts, use the quadratic formula:

${x}^{2} - x + 2 = 0 \setminus R i g h t a r r o w x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} = \frac{1}{2} \pm \frac{\sqrt{- 7}}{2}$

$= \frac{1}{2} \pm i \frac{\sqrt{7}}{2}$, which are (non-real) complex numbers. The graph of $f$ therefore has no $x$-intercepts (which can also be seen since the graph is upward pointing and the vertex is above the $x$-axis).