Graph $ \color{blue}{ f(x) = 28x^2} $

Step 1:

X - intercept is:

$$ \color{blue}{ x_1 = 0 } $$

To find the x-intercepts, we need to solve equation $ 28x^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,~0\right) $

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

$$ f(\color{blue}{0}) = 28 \cdot \color{blue}{0}^2 + 0 \cdot \color{blue}{0} + 0 = 0$$

Step 3:

Vertex is point: $V=\left(0,~0\right) $

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

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

In this example: $ a = 28, b = 0, c = 0 $. So, the x-coordinate of the vertex is:

$$ x = -\frac{b}{2a} = -\frac{ 0 }{ 2 \cdot 28 } = 0 $$$$ y = f \left( 0 \right) = 28 \left( 0 \right)^2 + 0 \cdot 0 ~ + ~ 0 = 0 $$

Step 4:

Focus is point: $ F=\left(0,~\dfrac{ 1 }{ 112 }\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