Question

1. y = -\frac{1}{5}(x - 5)^2-2

vertex:
axis of symmetry:
opens: up down
maximum minimum
max/min value:
y - intercept:

Explanation:

Step1: Identify the vertex - form of a parabola

The equation of a parabola in vertex - form is $y=a(x - h)^2+k$, where $(h,k)$ is the vertex. For the given equation $y =-\frac{1}{5}(x - 5)^2-2$, we have $h = 5$ and $k=-2$. So the vertex is $(5,-2)$.

Step2: Find the axis of symmetry

The axis of symmetry of a parabola in the form $y=a(x - h)^2+k$ is the vertical line $x = h$. Since $h = 5$, the axis of symmetry is $x = 5$.

Step3: Determine the direction of opening

If $a>0$, the parabola opens up; if $a<0$, the parabola opens down. Here, $a=-\frac{1}{5}<0$, so the parabola opens down.

Step4: Find the maximum/minimum value

Since the parabola opens down, it has a maximum value. The maximum value of the parabola is the $y$ - coordinate of the vertex. So the maximum value is $y=-2$.

Step5: Find the y - intercept

To find the y - intercept, set $x = 0$ in the equation $y=-\frac{1}{5}(x - 5)^2-2$.
\[

$$\begin{align*} y&=-\frac{1}{5}(0 - 5)^2-2\\ &=-\frac{1}{5}\times25-2\\ &=-5 - 2\\ &=-7 \end{align*}$$

\]

Answer:

Vertex: $(5,-2)$
Axis of symmetry: $x = 5$
Opens: down
Maximum value: $y=-2$
Max/Min value: Maximum
y - intercept: $(0,-7)$