Graph $ \color{blue}{ f(x) = -\dfrac{1}{10}x^2+\dfrac{6}{5}x+\dfrac{65}{2}} $
Solution
Step 1:
X - intercept are:
$$ \begin{aligned} & \color{blue}{ x_1 = -13 } \\[1 em] & \color{blue}{ x_2 = 25 } \end{aligned} $$To find the x-intercepts, we need to solve equation $ -\dfrac{1}{10}x^2+\dfrac{6}{5}x+\dfrac{65}{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,~\dfrac{ 65 }{ 2 }\right) $
To find y - coordinate of y - intercept, we need to compute $ f(0) $. In this example we have:
$$ f(\color{blue}{0}) = -\frac{ 1 }{ 10 } \cdot \color{blue}{0}^2 + \frac{ 6 }{ 5 } \cdot \color{blue}{0} + \frac{ 65 }{ 2 } = \frac{ 65 }{ 2 }$$Step 3:
Vertex is point: $V=\left(6,~\dfrac{ 361 }{ 10 }\right) $
To find the x - coordinate of the vertex we use formula:
$$ x = -\frac{b}{2a} $$In this example: $ a = -\frac{ 1 }{ 10 }, b = \frac{ 6 }{ 5 }, c = \frac{ 65 }{ 2 } $. So, the x-coordinate of the vertex is:
$$ x = -\frac{b}{2a} = -\frac{ \frac{ 6 }{ 5 } }{ 2 \cdot \left( -\frac{ 1 }{ 10 } \right) } = 6 $$$$ y = f \left( 6 \right) = -\frac{ 1 }{ 10 } \left( 6 \right)^2 + \frac{ 6 }{ 5 } \cdot 6 ~ + ~ \frac{ 65 }{ 2 } = \frac{ 361 }{ 10 } $$Step 4:
Focus is point: $ F=\left(6,~\dfrac{ 168 }{ 5 }\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 $