identify the vertex, axis of symmetry and concavity of each quadratic function.\nf(x)=2(x + 1)^2-8\nthe…

Explanation:

Step1: Recall vertex - form of quadratic function

The vertex - form of a quadratic function is $f(x)=a(x - h)^2+k$, where $(h,k)$ is the vertex and $x = h$ is the axis of symmetry. If $a>0$, the parabola opens upward (concave up), and if $a < 0$, the parabola opens downward (concave down).

Step2: Analyze $f(x)=2(x + 1)^2-8$

For $f(x)=2(x + 1)^2-8$, we have $a = 2$, $h=-1$, and $k=-8$. The vertex is $(-1,-8)$. The axis of symmetry is $x=-1$. Since $a = 2>0$, this function is concave up.

Step3: Analyze $f(x)=(x - 7)^2+5$

For $f(x)=(x - 7)^2+5$, we have $a = 1$, $h = 7$, and $k = 5$. The vertex is $(7,5)$. The axis of symmetry is $x = 7$. Since $a=1>0$, this function is concave up.

Step4: Analyze $f(x)=-(x + 10)^2$

For $f(x)=-(x + 10)^2$, we have $a=-1$, $h=-10$, and $k = 0$. The vertex is $(-10,0)$. The axis of symmetry is $x=-10$. Since $a=-1<0$, this function is concave down.

Answer:

For $f(x)=2(x + 1)^2-8$:

• The vertex is $(-1,-8)$
• The axis of symmetry is $x=-1$
• This function is concave up because $a = 2>0$

For $f(x)=(x - 7)^2+5$:

• The vertex is $(7,5)$
• The axis of symmetry is $x = 7$
• This function is concave up because $a = 1>0$

For $f(x)=-(x + 10)^2$:

• The vertex is $(-10,0)$
• The axis of symmetry is $x=-10$
• This function is concave down because $a=-1<0$