Answer:
Step-by-step explanation:
You have to put this into vertex form in order to find the time it was at its highest. To do this you have to complete the square. First rule is to set the equation equal to 0, then move the constant over to the other side of the equals sign. Let's do that:
[tex]-5t^2+20t=-105[/tex]
The next rule is that the leading coefficient HAS to be a positive 1, and right now ours is a -5. So we have to factor out the -5, leaving us:
[tex]-5(t^2-4t) =-105[/tex]
The next step is to take half the linear term, square it, and add it to both sides. Our linear term is a -4; half of that is -2, and squaring -2 gives you a 4. So adding in the 4 to the parenthesis on the left is a given; however, don't forget about the -5 hanging out in front as a multiplier. We didn't just add in a 4, we added in -5 times 4 which is -20. Putting all that together:
[tex]-5(t^2-4t+4)=-105-20[/tex]
The reason we do this is apparent on the left side. We create a perfect square binomial of the form
[tex]-5(t-2)^2=-135[/tex]
Notice that I also added the 2 negatives on the right at the same time. Now the last step is to move the -135 back over with the other guys and set it equal to y again:
[tex]-5(t-2)^2+135=y[/tex]
Now it's in the form
[tex]y=a(x-h)^2+k[/tex] where (h, k) are the vertex, or the highest point of the parabola: (2, 135). This means that at 2 seconds the object was at its highest point in its path of 135 m.
