Answer
A) Vertex: $(2.5,-3.1)$
B) Vertical intercept: $(0,-0.6)$
Horizontal intercepts: $(-0.284, 0),(5.284, 0)$
C) See the graph
D) Domain: All real numbers.
Range: $\left[-3.1, \infty\right) $
Work Step by Step
Given \begin{equation}
\begin{aligned}
p(x)&=\frac{2}{5} x^2-2 x-\frac{3}{5}\\
a& = \frac{2}{5} ,\quad b= -2, \quad c= -\frac{3}{5}.
\end{aligned}
\end{equation} Step 1: Determine whether the graph opens up or down. The parabola opens up when the constant $a$ is positive and opens down when $a$ negative. This parabola opens up since $a$ is positive.
Part A)
Step 2: Find the vertex. This function is in standard form for a quadratic, so we can use the formula: \begin{equation}
\begin{aligned}
& x=\frac{b}{2a}=-\frac{(-2)}{2\cdot \frac{2}{5}}=\frac{5}{2}=2.5 \\
& y=\frac{2}{5} \cdot 2.5^2-2(2.5)-\frac{3}{5}=-3.1.
\end{aligned}
\end{equation} The vertex of the function is $(2.5,-3.1)$.
Part B)
Step 3: Find the vertical and horizontal intercepts.
1) Find the vertical intercept which is equal to the constant $c$ if the parabola is in its standard form. \begin{equation}
\begin{aligned}
y& = -\frac{3}{5}=-0.6.
\end{aligned}
\end{equation} Vertical intercept: $(0,-0.6)$.
2) Find the horizontal intercept by setting the function to zero and solve. \begin{equation}
\begin{aligned}
\frac{2}{5} x^2-2 x-\frac{3}{5}&= 0\\
\left(\frac{2}{5} x^2-2 x-\frac{3}{5}\right)\cdot 5 & =0\cdot 5 \\
2 x^2-10 x-3 & =0\\
x&=\frac{-(-10) \pm \sqrt{(-10)^2-4(2)(-3)}}{2 \cdot 2}\\
x&=\frac{10 \pm \sqrt{124}}{4}\\
x&=2.5 \pm 2.784
\end{aligned}
\end{equation} \begin{equation}
\begin{aligned}
x & =2.5-2.784 \\
& =-0.284 \\
x & =2.5+2.784 \\
& =5.284.
\end{aligned}
\end{equation} Horizontal intercepts: $(-0.284, 0),(5.284, 0)$.
Part C) Sketch the graph as shown in the figure.
Part D) The domain and range of the function are given below:
Domain: All real numbers.
Range:$\left[-3.1, \infty\right)$
