Question

The function f(t)=-6t²+42t+20 models the approximate height of an object t seconds after it is launched. How high does the object go?

Answer 1

93.5 units

Step-by-step explanation:

The function that models the approximate height is given as:

`f\mleft(t\mright)=-6t^2+42t+20`

To determine how high the object goes, we find the maximum height ( or vertex) of the parabola.

First, find the equation of the line of symmetry using the formula below:

`x=-(b)/(2a)`
`\begin{gathered} a=-6,b=42 \\ \implies t=(-42)/(2(-6)) \\ =(42)/(12) \\ t=3.5 \end{gathered}`

Next, substitute t=3.5 into f(t) to find the maximum height.

`\begin{gathered} f\mleft(t\mright)=-6t^2+42t+20 \\ f(3.5)=-6(3.5)^2+42(3.5)+20 \\ =93.5 \end{gathered}`

The object goes as high as 93.5 units.

To demonstrate, the graph is attached here:

We see that the graph goes as high as 93.5 units (which was our result).