Graph $ \color{blue}{ f(x) = x^2+x+1} $

Step 1:

x - intercept does not exist.

To find the x-intercepts, we need to solve equation $ x^2+x+1 = 0 $. Since equation $ x^2+x+1 = 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,~1\right) $

To find y - coordinate of y - intercept, we need to compute $ f(0) $. In this example we have:

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

Step 3:

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

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

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

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

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

Step 4:

Focus is point: $ F=\left(-\dfrac{ 1 }{ 2 },~1\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