Question

graph each equation.

1. y = x^2 - 2x - 3
2. y=-x^2 - 6x - 10

identify the min/max value of each. then sketch the graph.

1. f(x)=-x^2 + 8x - 20
2. f(x)=-\frac{1}{3}x^2+\frac{4}{3}x-\frac{16}{3}

Explanation:

Step1: Recall vertex - form of a quadratic function

The vertex - form of a quadratic function is $y=a(x - h)^{2}+k$, where $(h,k)$ is the vertex of the parabola. For a quadratic function $y = ax^{2}+bx + c$, the $x$ - coordinate of the vertex is $h=-\frac{b}{2a}$, and then we substitute $x = h$ into the function to find $y = k$. If $a>0$, the parabola opens upward and has a minimum value. If $a < 0$, the parabola opens downward and has a maximum value.

Step2: Solve for problem 7

For the function $f(x)=-x^{2}+8x - 20$, where $a=-1$, $b = 8$, and $c=-20$.
First, find the $x$ - coordinate of the vertex: $h=-\frac{b}{2a}=-\frac{8}{2\times(-1)} = 4$.
Then, find the $y$ - coordinate of the vertex: $f(4)=-(4)^{2}+8\times4 - 20=-16 + 32-20=-4$.
Since $a=-1<0$, the parabola opens downward and the maximum value is $y=-4$.

Step3: Solve for problem 8

For the function $f(x)=-\frac{1}{3}x^{2}+\frac{4}{3}x-\frac{16}{3}$, where $a =-\frac{1}{3}$, $b=\frac{4}{3}$, and $c =-\frac{16}{3}$.
The $x$ - coordinate of the vertex is $h=-\frac{b}{2a}=-\frac{\frac{4}{3}}{2\times(-\frac{1}{3})}=2$.
The $y$ - coordinate of the vertex is $f(2)=-\frac{1}{3}(2)^{2}+\frac{4}{3}\times2-\frac{16}{3}=-\frac{4}{3}+\frac{8}{3}-\frac{16}{3}=-\frac{12}{3}=-4$.
Since $a=-\frac{1}{3}<0$, the parabola opens downward and the maximum value is $y = - 4$.

Answer:

1. Maximum value: $y=-4$.
2. Maximum value: $y=-4$.