Question

1. identify the maximum or minimum. then sta

a. $f(x)=x^{2}+2x - 8$ min $0:(-infty,infty)$ r e
b. $f(x)=x^{2}-6x + 14$ min $d(-infty,infty)$ l (
c. $f(x)=2x^{2}+4x - 6$ min $d(-infty,infty)$ r: -

1. simplify.

a. $sqrt{-49}=7i$
b. $6sqrt{-12}$

Explanation:

Step1: For part 6a - find the vertex of the quadratic function

For a quadratic function $f(x)=ax^{2}+bx + c$, the x - coordinate of the vertex is $x=-\frac{b}{2a}$. For $f(x)=x^{2}+2x - 8$, where $a = 1$, $b = 2$, and $c=-8$. Then $x=-\frac{2}{2\times1}=-1$. Substitute $x = - 1$ into $f(x)$: $f(-1)=(-1)^{2}+2\times(-1)-8=1 - 2 - 8=-9$. Since $a = 1>0$, the function has a minimum value of $-9$.

Step2: For part 6b - find the vertex of the quadratic function

For $f(x)=x^{2}-6x + 14$, with $a = 1$, $b=-6$, and $c = 14$. The x - coordinate of the vertex is $x=-\frac{-6}{2\times1}=3$. Substitute $x = 3$ into $f(x)$: $f(3)=3^{2}-6\times3 + 14=9-18 + 14=5$. Since $a = 1>0$, the function has a minimum value of $5$.

Step3: For part 6c - find the vertex of the quadratic function

For $f(x)=2x^{2}+4x - 6$, where $a = 2$, $b = 4$, and $c=-6$. The x - coordinate of the vertex is $x=-\frac{4}{2\times2}=-1$. Substitute $x=-1$ into $f(x)$: $f(-1)=2\times(-1)^{2}+4\times(-1)-6=2-4 - 6=-8$. Since $a = 2>0$, the function has a minimum value of $-8$.

Step4: For part 7b - simplify the square - root of a negative number

We know that $\sqrt{-1}=i$. So, $6\sqrt{-12}=6\sqrt{12\times(-1)}=6\sqrt{12}i=6\times2\sqrt{3}i = 12\sqrt{3}i$.

Answer:

6a. Minimum value is $-9$.
6b. Minimum value is $5$.
6c. Minimum value is $-8$.
7b. $12\sqrt{3}i$