please help! the approximate path of a projectile launched from 100 m above the ground is given by f(×)= -5×^2+40+100, where f(×) represents the height × seconds after it is released into the air. use the info to answer the questions below. 1. what is the factored form of the equation? 2 how long will it take the projectile to reach the ground?

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xKelvin xKelvin · Answer 1 · 2020-11-12T04:37:00+01:00

Answer:

Part 1) [tex]f(x)=-5(x-10)(x+2)[/tex]

Part 2) After 10 seconds.

Step-by-step explanation:

We have the function:

[tex]f(x)=-5x^2+40x+100[/tex]

Where f(x) represents the height x seconds after it is released into the air.

Part 1)

First, let's distribute a -5 out of the function:

[tex]f(x)=-5(x^2-8x-20)[/tex]

Now, let's factor. We want to pick two number that multiplies to -20 and adds to -8.

We can use -10 and 2. So:

[tex]f(x)=-5(x^2+2x-10x-20)[/tex]

Factor:

[tex]\begin{aligned}f(x)&=-5(x^2+2x-10x-20) \\ f(x)&=-5(x(x+2)-10(x+2)) \\ f(x)&=-5(x-10)(x+2) \end{aligned}[/tex]

Hence, our factored form is:

[tex]f(x)=-5(x-10)(x+2)[/tex]

Part 2)

We want to find after how many seconds does it take for the projectile to reach the ground.

If it reaches the ground, the height above the ground, f(x), will be 0.

So, we can set f(x) equal to 0 and solve for x. Instead of using the original equation, we can use the factored form. So:

[tex]0=-5(x-10)(x+2)[/tex]

Zero Product Property:

[tex]x-10=0\text{ or } x+2=0[/tex]

Solve for x:

[tex]x=10 \text{ or } x=-2[/tex]

Since x measures time, -2 does not make sense in this context.

Therefore, our solution is x=10.

So, after 10 seconds, the projectile will hit the ground.