Chapter 1.9, Problem 1MRPS

Solution

(a)

To find: The vertex form of the given function $y=-0.12x^2+1.2x+2$ .

The obtained vertex form is $y=-0.12{\left(x-5\right)}^2+5$ .

Given:

The equation is $y=-0.12x^2+1.2x+2$ .

Concept Used:

As the general form of a quadratic equation is $y=a{x}^2+bx+c$ . So the vertex form of the equation is defined as,

  $y=a{\left(x-b\right)}^2+c$

Where, $\left(b,c\right)$ is vertex of the equation.

Calculation:

Consider the model,

  $y=-0.12x^2+1.2x+2$ .

Make the perfect square of the right side of the equation.

  $\begin{array}{c}y=-0.12\left(x^2-10x\right)+2\\ =-0.12\left(x^2-2\cdot 5x+25-25\right)+2\\ =-0.12\left(x^2-2\cdot 5x+25\right)+2+\left(0.12\right)25\\ =-0.12{\left(x-5\right)}^2+5\end{array}$

Therefore, the obtained vertex form is $y=-0.12{\left(x-5\right)}^2+5$ .

(b)

To graph: The given equation $y=-0.12x^2+1.2x+2$ .

The obtained graph is as follows,

  

Graph:

Consider the given equation, $y=-0.12x^2+1.2x+2$ .

Now, make the table to graph the equation,

x	y	$y=-0.12x^2+1.2x+2$
0	2	$\left(0,2\right)$
5	5	$\left(5,5\right)$
$11.455$	0	$\left(11.455,0\right)$
$-1.455$	0	$\left(-1.455,0\right)$

Now, draw the graph.

  

Therefore, the above graph is the required graph.

(c)

To calculate: The maximum height of the given function $y=-0.12x^2+1.2x+2$ .

The maximum height is 5 feet.

Given Information:

Refer the information form the part (a) and part (b).

Calculation:

Consider the given information,

Refer the graph of the given equation from part (b).

  

As the given graph is a parabolic graph and it is opening to downward. So, the vertex will be the maximum height point of the equation. Thus, the maximum point is $\left(5,5\right)$ and the height is 5 feet.

Therefore, the maximum height is 5 feet.