Question

1. suppose a parabola has vertex (-2, -1) and also passes through the point (-4, -9). write the equation of the parabola in vertex form.
2. find the aos, vertex, max/min and value for the function f(x)=x² + 8

aos:____ vertex:____
max/min:______

Explanation:

Step1: Recall vertex - form of parabola

The vertex - form of a parabola is $y = a(x - h)^2+k$, where $(h,k)$ is the vertex of the parabola. Given the vertex $(-2,-1)$, we have $h=-2$ and $k = - 1$. So the equation of the parabola is $y=a(x + 2)^2-1$.

Step2: Find the value of $a$

Since the parabola passes through the point $(-4,-9)$, substitute $x=-4$ and $y = - 9$ into the equation $y=a(x + 2)^2-1$.
\[

$$\begin{align*} -9&=a(-4 + 2)^2-1\\ -9&=a(-2)^2-1\\ -9&=4a-1\\ 4a&=-9 + 1\\ 4a&=-8\\ a&=-2 \end{align*}$$

\]

Step3: Write the equation of the parabola

Substitute $a=-2$ into $y=a(x + 2)^2-1$. The equation of the parabola is $y=-2(x + 2)^2-1$.

Step4: Analyze the function $f(x)=x^{2}+8$

For a quadratic function in the form $y = ax^{2}+bx + c$, the axis of symmetry (AOS) is given by the formula $x=-\frac{b}{2a}$. For the function $f(x)=x^{2}+8$ (where $a = 1$, $b = 0$, $c = 8$), the axis of symmetry is $x=-\frac{0}{2\times1}=0$.

Step5: Find the vertex

The vertex of a quadratic function $y = ax^{2}+bx + c$ has $x$ - coordinate $x=-\frac{b}{2a}$ and $y$ - coordinate $y=f(-\frac{b}{2a})$. Since $x = 0$, $y=f(0)=0^{2}+8=8$. So the vertex is $(0,8)$.

Step6: Determine the maximum or minimum

Since $a = 1>0$, the parabola opens upward. So the function has a minimum value. The minimum value of the function is $y = 8$ (the $y$ - coordinate of the vertex).

Answer:

1. $y=-2(x + 2)^2-1$
2. AOS: $x = 0$

Vertex: $(0,8)$
Max/Min: Minimum, value is $8$