Graph $ \color{blue}{ f(x) = x^2-3x+2} $

Step 1:

X - intercept are:

$$ \begin{aligned} & \color{blue}{ x_1 = 1 } \\[1 em] & \color{blue}{ x_2 = 2 } \end{aligned} $$

To find the x-intercepts, we need to solve equation $ x^2-3x+2 = 0 $. (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:

$$ f(\color{blue}{0}) = 1 \cdot \color{blue}{0}^2 -3 \cdot \color{blue}{0} + 2 = 2$$

Step 3:

Vertex is point: $V=\left(\dfrac{ 3 }{ 2 },~-\dfrac{ 1 }{ 4 }\right) $

To find the x - coordinate of the vertex we use formula:

$$ x = -\frac{b}{2a} $$

In this example: $ a = 1, b = -3, c = 2 $. So, the x-coordinate of the vertex is:

$$ x = -\frac{b}{2a} = -\frac{ -3 }{ 2 \cdot 1 } = \frac{ 3 }{ 2 } $$$$ y = f \left( \frac{ 3 }{ 2 } \right) = 1 \left( \frac{ 3 }{ 2 } \right)^2 - 3 \cdot \frac{ 3 }{ 2 } ~ + ~ 2 = -\frac{ 1 }{ 4 } $$

Step 4:

Focus is point: $ F=\left(\dfrac{ 3 }{ 2 },~0\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 $

Quadratic Function Grapher