Answer:
[tex]Vertex = (\frac{1}{2},\frac{7}{4})[/tex]
Step-by-step explanation:
Given
[tex]f(x) = x^2 - x +2[/tex]
Required
The vertex
We have:
[tex]f(x) = x^2 - x +2[/tex]
First, we express the equation as:
[tex]f(x) = a(x - h)^2 +k[/tex]
Where
[tex]Vertex = (h,k)[/tex]
So, we have:
[tex]f(x) = x^2 - x +2[/tex]
--------------------------------------------
Take the coefficient of x: -1
Divide by 2: (-1/2)
Square: (-1/2)^2
Add and subtract this to the equation
--------------------------------------------
[tex]f(x) = x^2 - x +2[/tex]
[tex]f(x) = x^2 - x + (-\frac{1}{2})^2+2 -(-\frac{1}{2})^2[/tex]
[tex]f(x) = x^2 - x + \frac{1}{4}+2 -\frac{1}{4}[/tex]
Expand
[tex]f(x) = x^2 - \frac{1}{2}x- \frac{1}{2}x + \frac{1}{4}+2 -\frac{1}{4}[/tex]
Factorize
[tex]f(x) = x(x - \frac{1}{2})- \frac{1}{2}(x - \frac{1}{2})+2 -\frac{1}{4}[/tex]
Factor out x - 1/2
[tex]f(x) = (x - \frac{1}{2})(x - \frac{1}{2})+2 -\frac{1}{4}[/tex]
[tex]f(x) = (x - \frac{1}{2})^2+2 -\frac{1}{4}[/tex]
[tex]f(x) = (x - \frac{1}{2})^2+ \frac{8 -1 }{4}[/tex]
[tex]f(x) = (x - \frac{1}{2})^2+ \frac{7}{4}[/tex]
Compare to: [tex]f(x) = a(x - h)^2 +k[/tex]
[tex]h = \frac{1}{2}[/tex]
[tex]k = \frac{7}{4}[/tex]
Hence:
[tex]Vertex = (\frac{1}{2},\frac{7}{4})[/tex]
