Chapter 4.7, Problem 20PPS

To determine

To graph: the quadratic equation as mentioned in the problem.

Explanation of Solution

Given information:

The quadratic equation is $h=-4.9{\left(t-4\right)}^2+80$ .

Graph:

The graph mentioned below is in real number line of x axis is $t$ and y axis is $h$ .

Where vertexY $V\left(4,80\right)$ .

  

Interpretation:

Consider the quadratic equation13 $h=-4.9{\left(t-4\right)}^2+80$ .

The mentioned quadratic equation above is in vertex form

  $y=a{\left(x-h\right)}^2+k$ where $\left(h,k\right)$ is the vertex.

Therefore the vertex of equation is $V\left(4,80\right)$ .

We can write the mentioned quadratic equation in the standard form as mentioned below

  $\begin{array}{l}h=-4.9{\left(t-4\right)}^2+80\Rightarrow h=-4.9\left(t^2+16-8t\right)+80\Rightarrow h=-4.9t^2-78.4+39.2t+80\\ \Rightarrow h=-4.9t^2+39.2t+\mathrm{1.6.}\end{array}$ .

Which is comparable to $a{x}^2+bx+c$ where $a=-4.9$ which is less than $0$ .

Since $a<0$ the direction of graph will be negative side of y axis.

The discriminant of standard form of quadratic equation $h=-4.9{\left(t-4\right)}^2+80$ is $D=b^2-4*a*c\Rightarrow d={\left(39.2\right)}^2-4*\left(-4.9\right)*\left(1.6\right)=1536.64+31.36=1568$ .

Here $D>0$ .

Therefore the graph will lie on above x axis.

At $t=0$ the $h=-4.9{\left(t-4\right)}^2+80$ will yield $h=-78.4$ .

Therefore graph will cut h axis at $\left(0,-78.4\right)$ .

Therefore the graph will be as mentioned below.

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