Graph $ \color{blue}{ f(x) = 2x^2-11x-6} $

Step 1:

X - intercept are:

$$ \begin{aligned} & \color{blue}{ x_1 = -\frac{ 1 }{ 2 } } \\[1 em] & \color{blue}{ x_2 = 6 } \end{aligned} $$

To find the x-intercepts, we need to solve equation $ 2x^2-11x-6 = 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,~-6\right) $

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

$$ f(\color{blue}{0}) = 2 \cdot \color{blue}{0}^2 -11 \cdot \color{blue}{0} -6 = -6$$

Step 3:

Vertex is point: $V=\left(\dfrac{ 11 }{ 4 },~-\dfrac{ 169 }{ 8 }\right) $

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

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

In this example: $ a = 2, b = -11, c = -6 $. So, the x-coordinate of the vertex is:

$$ x = -\frac{b}{2a} = -\frac{ -11 }{ 2 \cdot 2 } = \frac{ 11 }{ 4 } $$$$ y = f \left( \frac{ 11 }{ 4 } \right) = 2 \left( \frac{ 11 }{ 4 } \right)^2 - 11 \cdot \frac{ 11 }{ 4 } ~ - ~ 6 = -\frac{ 169 }{ 8 } $$

Step 4:

Focus is point: $ F=\left(\dfrac{ 11 }{ 4 },~-21\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