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Andrei [34K]
3 years ago
10

1.) What is the equation of the path of firework #1? Write your equation in general form.

Mathematics
1 answer:
valina [46]3 years ago
3 0

1. I'm assumig that the paths are perfect parabolas

this means that their general forms can be written in y=ax^2+bx+c

it's easier to find vertex form first then expand to get general form

vertex form is y=a(x-h)^2+k where the vertex is (h,k) and a is a constant


firework #1

vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

h_1=a(t-10)^2+50

to find the value of a, subsitute another point

(0,0)

0=a(0-10)^2+50)

0=100a+50

a=\frac{-1}{2}

so the equation in vertex form is h_1=\frac{-1}{2}(t-10)^2+50

expand to get general form

h_1=\frac{-1}{2}(t^2-20t+100)+50

h_1=\frac{-1}{2}t^2+10t-50+50

h_1=\frac{-1}{2}t^2+10t





2.

same as last time

vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72

equation is h_2=a(t-10)^2+72

find a

use another point

(0,22)

22=a(0-10)^2+72

22=100a+72

-50=100a

a=\frac{-1}{2}

so the equation in vertex form is h_2=\frac{-1}{2}(t-10)^2+72




3.

range is the numbers that h is allowed to be

think about what h represents. it represents the height of the rocket

from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd

so range is 0 to 50 or 0≤h≤50


domain is the numbers that t is allowed to be

think about what t represents. it represents how long the rocket has been flying

it will stop flying when it hits the ground or at t=20

it starts flying at t=0

so domain is from 0 to 20 or 0≤t≤20

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\huge \boxed{\mathfrak{Question} \downarrow}

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