44.5k views
1 vote
The general form of an parabola is 4y2+40y+3x+103=0. What is the standard form of the parabola?

User Mato
by
5.3k points

2 Answers

7 votes

Answer:

The answer is
(y+5)^(2) =-(3)/(4) (x+1)

Explanation:

I actually don't know the step by steps myself, so I just searched on

The general form of an parabola is 4y2+40y+3x+103=0. What is the standard form of-example-1
User Taga
by
5.3k points
3 votes

9514 1404 393

Answer:

x = -4/3y² -40/3y -103/3

x = -4/3(y +5)² -1

Explanation:

Solve for x.

x = (4y² +40y +103)/(-3)

x = -4/3y² -40/3y -103/3 . . . . 'standard form' in the US

__

Taking our clues from the graph*, we can write the vertex form equation as ...

x = -4/3(y +5)² -1 . . . . . . 'standard form' in other places

_____

* The vertex is (-1, -5), so for some leading coefficient, the equation will be ...

x = a(y -(-5))² +(-1) = a(y +5)² -1

The value of 'a' is the scale factor. Here, that is the difference between the parabola value (x = -2 1/3) and the vertex value (x = -1) one unit away from the vertex.

The general form of an parabola is 4y2+40y+3x+103=0. What is the standard form of-example-1
User OldSchool
by
6.4k points