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3 years ago

12. Write the equation of a parabola with vertex at (2, -3) and focus at (2,5),

Mathematics

1 answer:

Tatiana [17]3 years ago

7 0

Step-by-step explanation:

The basic form of equation:

(x-h)²=4a(y-k),

(h,k)=coordinates of vertex

(h, k+a) = coordinate of focus

For given parabola:

axis of symmetry: x=2

(h, k) =(2,-3)

(h, k+a)=(2,5)

k+a=5

-3+a=5

a=8(distance from vertex to focus on the axis of symmetry)

equation: (x-2)²=4×8(y+3)

(x-2)²= 32(y+3)

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Write the equation using function notation f(x). then find f(0) 2x^2 - 3y = 6

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$\quad \huge \quad \quad \boxed{ \tt \:Answer }$

$\qquad \tt \rightarrow \: f(0) = -2$

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$\large \tt Solution \: :$

$\qquad \tt \rightarrow \: 2 {x}^{2} - 3y = 6$

$\qquad \tt \rightarrow \: 3y + 6 = 2 {x}^{2}$

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[ here, y can be replaced with f(x) because y is a function of x ]

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Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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Read 2 more answers

If M is the set of all square of integers that are less than 100 and N is the set of all positive odd numbers that are under 15,

ollegr [7]

A set is simply the group of numbers.

The set of M and N are: $M = \{1,4,9,16,25,36,49,64,81\}$ and $N= \{1,3,5,7,9,11,13\}$ .
The set of M n N is: $M\ n\ N = \{1,9\}$ .
The set of M u N is: $M\ u\ N = \{1,3,4,5,7,9,11,13,16,25,36,49,64,81\}$

(a) The set of M and N

Given that:

M = squares of integers less than 100

N = Positive odd numbers less than 15

So, we have:

$M = \{1,4,9,16,25,36,49,64,81\}$

$N= \{1,3,5,7,9,11,13\}$

(b) M n N and M u N

M n N is the set of common elements between sets M and N.

So, we have:

$M\ n\ N = \{1,9\}$

M u N is the set of all elements in sets M and N.

So, we have:

$M\ u\ N = \{1,3,4,5,7,9,11,13,16,25,36,49,64,81\}$