An object is launched from a platform. its height in meters X seconds after the launch is modeled by h(x)=-5x^2+20x+60 how many seconds after launch Will the object land on the ground

6 seconds after launch.

Step-by-step explanation:

You should make sure  there aren't odd symbols in your question when you post it.

Assuming it is just the equation -5x^2 + 20x + 60, if you graph it you can see the path of the ball.  at x=0 this is the starting point It then hits the ground when it touches the x axis.  So this is finding zeroes of a quadratic.

You could do this several ways; find a way to factor it, guess and check, the quadratic formula or completing the square.  I will complete the square as that is what I am most used to.  If you would like to see another let me know.

The first step in completing the square is making sure the x^2 has a coefficient of 1, so now it has -5 so we have to factor out -5 from the equation.

-5x^2+20x+60 = -5(x^2 - 4x - 12)

Now you only really need to focus on the factored expression, but I will continue writing everything.  The next step is to find (b/2)^2 where b is the coefficient of the x term.  In this case -4.  so (b/2)^2 = (-4/2)^2 = 4.

Again, focusing on x^2 - 4x - 12 you want to add and subtract (b/2)^2 from that expression.  since you are adding and subtracting you are not changing the value, but we can use this.

-5(x^2 - 4x - 12) = -5(x^2 - 4x - 12 + 4 - 4)

Here you want to rearrange it a bit.  Originally c was the constant term, -12, now you want it to be the positive (b/2)^2, which is positive 4. also you can combine the two other constant terms.  the original c and -(b/2)^2, but I will hold off for now so I don't do too much at once.

-5(x^2 - 4x - 12 + 4 - 4) = -5(x^2 - 4x + 4 - 12 - 4)

Now you focus on x^2 - 4x + 4.  Hope fully you recognize this is the same as (x-2)^2.  This always happens at this step of the process.  also notice -2 = b/2.  Using only variables here are the first steps.

ax^2 + bx + c

a(x^2 + (b/a)x + (c/a))

a(x^2 + (b/a)x + (c/a) + (b/(2a))^2 - (b/(2a)^2)

a(x^2 + (b/a)x + (b/(2a))^2 + (c/a) - (b/(2a))^2)

a((x + b/(2a))^2 + (c/a) - (b/(2a))^2)

Again, being able to make x^2 - 4x + 4 into (x-2)^2 or x^2 + (b/a)x + (b/(2a))^2 into (x + b/(2a))^2 willa lways happen.  this is because if you expand (x + b/(2a))^2 you always get x^2 + (b/a)x + (b/(2a))^2.

Now I would combine the -12-4

-5(x^2 - 4x + 4 - 12 - 4) = -5((x - 2)^2 - 12 - 4) = -5((x - 2)^2 - 16)

If you redistribute the -5 you get vertex form, but I am going to stop here  because I will just undo that in the next step.  So this is the form you want.  Finally we can find when it equals 0.  So, you set this equal to 0 and use algebra to solve.

-5((x - 2)^2 - 16) = 0

Divide both sides by -5

(x-2)^2 - 16 = 0

add 16 to both sides

(x-2)^2 = 16

Take the square root of both sides, but also count botht he positive and negative version of the answer.  the reason is both 2 and -2 squared get you 4.

x - 2 = +/-4

I am using +/- to indicate I am using both positive and negative 4.  Now though add 2 to both sides.  since you have +/- 4 you are going to get two different results.  -4+2 and 4+2

x = -2 and 6.

now, the question wants times after x=0 (the start) so you only get 6.  so x=6, or in other words 6 seconds after launch.

Let me know if you have any questions.

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step-by-step explanation:

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