Question

1. $f(x)=3(x + 2)^2+8$ circle one: vertex or standard vertex min or max axis of symmetry y - intercept direction of opening

Explanation:

Step1: Identify the form

The function $f(x)=3(x + 2)^2+8$ is in vertex - form $y=a(x - h)^2+k$, where $(h,k)$ is the vertex. Here $h=-2$ and $k = 8$, so the vertex is $(-2,8)$.

Step2: Determine min/max

Since $a = 3>0$, the parabola opens upward and has a minimum value. The minimum value is $y = 8$ (the $y$-coordinate of the vertex).

Step3: Find axis of symmetry

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

Step4: Calculate y - intercept

Set $x = 0$ in the function: $f(0)=3(0 + 2)^2+8=3\times4+8=12 + 8=20$. So the $y$-intercept is $20$.

Step5: Determine opening direction

Since $a=3>0$, the parabola opens upward.

Answer:

Property	Value
Min or Max	Min, $y = 8$
Axis of Symmetry	$x=-2$
y - intercept	$20$
Direction of Opening	Upward