Graph $ \color{blue}{ f(x) = \dfrac{1}{2}x^2-x+2} $
Step 1:
x - intercept does not exist.
To find the x-intercepts, we need to solve equation $ \dfrac{1}{2}x^2-x+2 = 0 $. Since equation $ \dfrac{1}{2}x^2-x+2 = 0 $ does not have real soultions we conclude that the function does not have x - intercepts too. (use the quadratic equation solver to view a detailed explanation of how to solve the equation)
Step 2:
Y - intercept is point: $ y-inter=\left(0,~2\right) $
To find y - coordinate of y - intercept, we need to compute $ f(0) $. In this example we have:
$$ f(\color{blue}{0}) = \frac{ 1 }{ 2 } \cdot \color{blue}{0}^2 -1 \cdot \color{blue}{0} + 2 = 2$$Step 3:
Vertex is point: $V=\left(1,~\dfrac{ 3 }{ 2 }\right) $
To find the x - coordinate of the vertex we use formula:
$$ x = -\frac{b}{2a} $$In this example: $ a = \frac{ 1 }{ 2 }, b = -1, c = 2 $. So, the x-coordinate of the vertex is:
$$ x = -\frac{b}{2a} = -\frac{ -1 }{ 2 \cdot \frac{ 1 }{ 2 } } = 1 $$$$ y = f \left( 1 \right) = \frac{ 1 }{ 2 } \left( 1 \right)^2 - 1 \cdot 1 ~ + ~ 2 = \frac{ 3 }{ 2 } $$Step 4:
Focus is point: $ F=\left(1,~2\right)$
The x - coordinate of the focus is $ x = -\dfrac{b}{2a} $
The y - coordinate of the focus is $ y = \dfrac{1-b^2}{4a} + c $