Question

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

Answer 1

If `y=f(x)=x^2-x+2`, the `y`-intercept is `f(0)=2`, the vertex can be found by completing the square (shown below) to be at the point `(1/2,f(1/2))=(1/2,7/4)`, 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(x)=x^2-x+2=x^2-x+(-1/2)^2+2-(-1/2)^2`

`=(x-1/2)^2+2-1/4=(x-1/2)^2+7/4`.

This shows that the vertex, or low point of this upward pointing parabola, is at `(x,y)=(1/2,f(1/2))=(1/2,7/4)`.

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

`x^2-x+2=0 \Rightarrow x=(1 pm sqrt(1-4*1*2))/(2*1)=1/2 pm sqrt(-7)/2`

`=1/2 pm i 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).