Chapter 9.3, Problem 12OE

To determine

a)

To find: The vertex of parabola.

Answer to Problem 12OE

The vertex of the given parabola is $\left(4,-4\right)$ .

Explanation of Solution

Given information: The equation of parabola $y+4=-\frac{1}{36}{\left(x-4\right)}^2$

Formula used: $y=\frac{1}{4c}{\left(x-h\right)}^2+k$

Calculation: Rearranging the given equation,

  $\begin{array}{l}y+4=-\frac{1}{36}{\left(x-4\right)}^2\\ y=-\frac{1}{36}{\left(x-4\right)}^2-4\end{array}$

Compare $y=-\frac{1}{36}{\left(x-4\right)}^2-4$ to $y=\frac{1}{4c}{\left(x-h\right)}^2+k$

Hence, the vertex $\left(h,k\right)=\left(4,-4\right)$ .

Therefore, the required vertex is $\left(4,-4\right)$ .

To determine

b)

To find: The value of $\left|c\right|$ .

Answer to Problem 12OE

The value of $\left|c\right|$ is $9$ .

Explanation of Solution

Given information: The equation of parabola $y+4=-\frac{1}{36}{\left(x-4\right)}^2$

Formula used: $y=a{\left(x-h\right)}^2+k$

Calculation: Rearranging the given equation,

  $\begin{array}{l}y+4=-\frac{1}{36}{\left(x-4\right)}^2\\ y=-\frac{1}{36}{\left(x-4\right)}^2-4\end{array}$

Compare $y=-\frac{1}{36}{\left(x-4\right)}^2-4$ to $y=a{\left(x-h\right)}^2+k$

Hence,

  $\begin{array}{l}\because a=\left|-\frac{1}{36}\right|\&\left|c\right|=\left|\frac{1}{4a}\right|\\ \therefore \left|c\right|=\left|\frac{1}{4\times \left(-\frac{1}{36}\right)}\right|\\ =\left|\frac{1}{-\frac{4}{36}}\right|\end{array}$

  $\begin{array}{l}=\left|\frac{1}{-\frac{1}{9}}\right|\\ =\left|-9\right|\\ \therefore \left|c\right|=9\end{array}$

Therefore, the value of $\left|c\right|$ is $9$ .

To determine

c)

To find: The opening of parabola.

Answer to Problem 12OE

The parabola will open downwards.

Explanation of Solution

Given information: The equation of parabola $y+4=-\frac{1}{36}{\left(x-4\right)}^2$

Formula used: $y=a{\left(x-h\right)}^2+k$

Calculation: Compare $y=-\frac{1}{36}{\left(x-4\right)}^2-4$ to $y=a{\left(x-h\right)}^2+k$

Hence,

  $\begin{array}{l}\therefore a=\frac{-1}{36}\\ \therefore a<0\end{array}$

Therefore, the parabola will open downwards.