Graph $ \color{blue}{ f(x) = -x^2+6x+29} $

Step 1:

X - intercept are:

$$ \begin{aligned} & \color{blue}{ x_1 = 3-\sqrt{ 38 } } \\[1 em] & \color{blue}{ x_2 = 3+\sqrt{ 38 } } \end{aligned} $$

To find the x-intercepts, we need to solve equation $ -x^2+6x+29 = 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,~29\right) $

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

$$ f(\color{blue}{0}) = -1 \cdot \color{blue}{0}^2 + 6 \cdot \color{blue}{0} + 29 = 29$$

Step 3:

Vertex is point: $V=\left(3,~38\right) $

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

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

In this example: $ a = -1, b = 6, c = 29 $. So, the x-coordinate of the vertex is:

$$ x = -\frac{b}{2a} = -\frac{ 6 }{ 2 \cdot \left( -1 \right) } = 3 $$$$ y = f \left( 3 \right) = -1 \left( 3 \right)^2 + 6 \cdot 3 ~ + ~ 29 = 38 $$

Step 4:

Focus is point: $ F=\left(3,~\dfrac{ 151 }{ 4 }\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