Answer
1) Vertex :$(3,2.5)$
2) Vertical intercept: $(0,-21.8)$
Horizontal intercepts:$\left(2.038, 0\right),\left(3.962,0\right)$
3) Domain: All real numbers.
Range: $(-\infty, 2.5] $
Work Step by Step
Given \begin{equation}
\begin{aligned}
f(x)&=-2.7 x^2+16.2 x-21.8.
\end{aligned}
\end{equation} We determine the constants $a$, $b$, $c$: $$\begin{align*}
f(x)&=ax^2+bx+c\\
a& = -2.7, b= 16.2, c= -21.8.
\end{align*}$$ 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 down since $a$ is negative.
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}{2 a} \\
& =\frac{-( 16.2)}{2(-2.7)} \\
& =3\\
f(3) & =-2.7\cdot (3)^2+16.2\cdot (3)-21.8 \\
& =2.5.
\end{aligned}
\end{equation} The vertex of the function is $(3,2.5)$.
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& = -21.8.
\end{aligned}
\end{equation} Vertical intercept: $(0,-21.8)$.
2) Find the horizontal intercept by setting the function to zero and solve.
\begin{equation}
\begin{aligned}
\left(-2.7 x^2+16.2 x-21.8 \right)\cdot (-10)& =0\cdot (-10) \\
27 x^2-162 x+218 & =0 \\
x & =\frac{-(-162)\pm \sqrt{(-162)-4\cdot(27)\cdot(218)}}{2\cdot(27)} \\
& = \frac{162\pm \sqrt{2700}}{54} \\
x & =3 \pm 0.9623\\
x_1& = 3-0.9623= 2.038\\
x_2& = 3+0.9623= 3.962.
\end{aligned}
\end{equation} The graph of the function is shown in the figure with the required points shown.
