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Firlakuza [10]
3 years ago
13

If the numbers listed below represent temperatures in degrees Celsius, which is the coldest?

Mathematics
2 answers:
MAXImum [283]2 years ago
8 0
-3 is what I think it is
Aloiza [94]2 years ago
5 0

Answer:

-3 would be the coldest. Colder room temperatures fall below 0, and -3 is the lowest number

Step-by-step explanation:

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NO LINKS!! Please help me with these graphs. Part 1​
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Problem 9

The instructions aren't stated anywhere, but I'm assuming your teacher wants you to find the equation of each parabola.

The vertex here is (h,k) = (-6,-4) which you have correctly determined.

This means

y = a(x-h)^2 + k\\\\y = a(x-(-6))^2 +(-4)\\\\y = a(x+6)^2 - 4\\\\

Next we plug in one of the other points on the parabola. We cannot pick the vertex again. Let's pick the point (-4,0) which is one of the x intercepts. We'll do this to solve for 'a'

y = a(x+6)^2 - 4\\\\0 = a(-4+6)^2 - 4\\\\0 = a(2)^2 - 4\\\\0 = a(4) - 4\\\\0 = 4a-4\\\\4a-4 = 0\\\\4a = 4\\\\a = 4/4\\\\a = 1\\\\

This means

y = a(x+6)^2 - 4\\\\y = 1(x+6)^2 - 4\\\\y = (x+6)^2 - 4\\\\

represents the equation of the parabola in vertex form.

<h3>Answer: y = (x+6)^2 - 4\\\\</h3>

========================================================

Problem 10

The vertex is (h,k) = (-2,-6)

So,

y = a(x-h)^2 + k\\\\y = a(x-(-2))^2 + (-6)\\\\y = a(x+2)^2 - 6\\\\

Now plug in another point on the parabola like (-1,-8) and solve for 'a'

y = a(x+2)^2 - 6\\\\-8 = a(-1+2)^2 - 6\\\\-8 = a(1)^2 - 6\\\\-8 = a(1) - 6\\\\a-6 = -8\\\\a = -8+6\\\\a = -2\\\\

<h3>Answer:  y = -2(x+2)^2 - 6\\\\</h3>

For each equation, you could optionally expand things out to get it into y = ax^2+bx+c form, but I think it's fine to leave it as vertex form.

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