Chapter 9, Problem 14MCQ

To determine

Describe how the graph of each function is related to the graph of $h=t^2\text{and}h=-16t^2+14$

Answer to Problem 14MCQ

The graph of $h=t^2\text{and}h=-16t^2+14$ is the graph of $h=t^2$ reflected across the x − axis and vertically stretched by a factor 16 and vertically translated 14 units up.

Explanation of Solution

Given: The functions $h=t^2\text{and}h=-16t^2+14$

Concept Used:

The functions $h=t^2\text{and}h=-16t^2+14$

  $\begin{array}{l}\text{Vertical}\text{Stretch}\text{and}\text{Vertical}\text{Compression:}\\ \text{y}\text{=}\text{a}\text{f}\text{(x)}\text{,}\text{a}\text{>1}\text{,}\text{Srtetch}\text{the}\text{graph}\text{of}\text{f(x)}\text{vertically}\text{by}\text{a}\text{factor}\text{of}\text{a}\text{.}\\ \text{y}\text{=}\text{a}\text{f}\text{(x)}\text{,}\text{0}\text{<}\text{a}\text{<1}\text{,}\text{Compress}\text{the}\text{graph}\text{of}\text{f(x)}\text{vertically}\text{by}\text{a}\text{factor}\text{of}\text{a}\end{array}$

The graph of $f(x)=a{x}^2+c$ stretches or compress the parent graph vertically The coefficient of $x^2$ term is negative, so the graph is reflected across the x − axis. If c is positive, the graph translated up and if c is negative, the graph translated down from the parent graph. Since $a=16$, the graph is vertically stretched by a factor 16.   $c=14$, the graph is translated 100 units up from the graph $h=t^2$ So, the graph of $h=-16t^2+14$ is the graph of $h=t^2$ reflected across the x − axis and vertically stretched by a factor 16 and vertically translated 14 units up.

Thus, the graph of $h=-16t^2+14$ is the graph of $h=t^2$ reflected across the x − axis and vertically stretched by a factor 16 and vertically translated 14 units up.