Question

2.3 writing quadratic functions (pp. 141 - 150) learning target: write equations of quadratic functions using given characteristics. the table or graph represents a quadratic function. write an equation of the function in standard form. 17.

x	- 5	- 4	- 3	- 2
g(x)	5	2	5	14	18.

1. vertex: (-1, 8); passes through (0, 2) 20. passes through (-2, 0), (4, 18), and (10, 0) 21. a researcher models the average height of the waves on a beach during a storm using a parabola, where x is the number of hours after the storm began and y is the average height (in feet) of the waves. a. write a function that models the average height over time. what is the tallest average wave height? b. what is the average rate of change in height over the interval in which the function is decreasing? increasing? compare the average rates of change.

Explanation:

Step1: Recall quadratic - function standard form

The standard form of a quadratic function is $y = ax^{2}+bx + c$.

Step2: Solve problem 19

The vertex - form of a quadratic function is $y=a(x - h)^{2}+k$, where $(h,k)$ is the vertex. Given the vertex $(-1,8)$, the function is $y=a(x + 1)^{2}+8$. Substitute the point $(0,2)$ into the equation:
\[

$$\begin{align*} 2&=a(0 + 1)^{2}+8\\ 2&=a+8\\ a&=-6 \end{align*}$$

\]
Expand $y=-6(x + 1)^{2}+8$:
\[

$$\begin{align*} y&=-6(x^{2}+2x + 1)+8\\ y&=-6x^{2}-12x-6 + 8\\ y&=-6x^{2}-12x + 2 \end{align*}$$

\]

Step3: Solve problem 20

Since the quadratic function passes through $(-2,0)$ and $(10,0)$, the equation can be written in factored form as $y=a(x + 2)(x - 10)$. Substitute the point $(4,18)$ into the equation:
\[

$$\begin{align*} 18&=a(4 + 2)(4 - 10)\\ 18&=a(6)(-6)\\ 18&=-36a\\ a&=-\frac{1}{2} \end{align*}$$

\]
Expand $y=-\frac{1}{2}(x + 2)(x - 10)$:
\[

$$\begin{align*} y&=-\frac{1}{2}(x^{2}-10x+2x - 20)\\ y&=-\frac{1}{2}(x^{2}-8x - 20)\\ y&=-\frac{1}{2}x^{2}+4x + 10 \end{align*}$$

\]

Step4: Solve problem 21a

Since the vertex of the parabola is $(8,22)$ and it passes through $(0,6)$ and $(16,6)$, the vertex - form of the quadratic function is $y=a(x - 8)^{2}+22$. Substitute $(0,6)$ into the equation:
\[

$$\begin{align*} 6&=a(0 - 8)^{2}+22\\ 6&=64a+22\\ 64a&=-16\\ a&=-\frac{1}{4} \end{align*}$$

\]
The function is $y =-\frac{1}{4}(x - 8)^{2}+22=-\frac{1}{4}(x^{2}-16x + 64)+22=-\frac{1}{4}x^{2}+4x + 6$. The tallest average wave height is the $y$ - value of the vertex, which is $22$ feet.

Step5: Solve problem 21b

The function $y =-\frac{1}{4}x^{2}+4x + 6$ is increasing on the interval $0\leq x\leq8$ and decreasing on the interval $8\leq x\leq16$.
The average rate of change on the increasing interval $[0,8]$ is $\frac{y(8)-y(0)}{8 - 0}=\frac{22 - 6}{8}=\frac{16}{8}=2$.
The average rate of change on the decreasing interval $[8,16]$ is $\frac{y(16)-y(8)}{16 - 8}=\frac{6 - 22}{8}=\frac{-16}{8}=-2$. The magnitude of the average rates of change is the same, but the signs are opposite.

Answer:

1. $y=-6x^{2}-12x + 2$
2. $y=-\frac{1}{2}x^{2}+4x + 10$

21a. $y =-\frac{1}{4}x^{2}+4x + 6$, tallest height is 22 feet
21b. Increasing interval average rate of change: 2, Decreasing interval average rate of change: - 2, magnitudes are the same, signs are opposite.