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2 years ago

Which equation describes a parabola that opens left or right and whose vertex is at the point (h, v)?

Mathematics

2 answers:

svetlana [45]2 years ago

7 0

The <em><u>correct answer</u></em> is:

x = a(y-v)²+h

Explanation:

Vertex form of a parabola differs depending on the direction the parabola opens. If it opens up or down, the equation is y = a(x-h)²+v, where (h, v) is the vertex.

If, however, the parabola opens left or right, the equation is x = a(y-v)²+h, where (h, v) is the vertex.

nalin [4]2 years ago

4 0

Vertex is at the point ( h , v )
The equation of a parabola is:
x = a ( y - v )² + h
Answer: B )

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Choose the solution(s) of the following system of equations x2 + y2 = 6 x2 – y = 6

ElenaW [278]

<h2>Answer</h2>

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

Or as ordered pairs: $(\sqrt{6} ,0),(-\sqrt{6} ,0),(\sqrt{5} ,-1),(-\sqrt{5} ,-1)$

<h2>Explanation</h2>

Lets solve our system of equations step by step

$x^2+y^2=6$ equation (1)

$x^2-y=6$ equation (2)

1. Solve for $x^2$ in equation (2)

$x^2-y=6$

$x^2=6+y$ equation (3)

2. Replace equation (3) in equation (1) and solve for $y$

$x^2+y^2=6$

$6+y+y^2=6$

$y^2+y=0$

$y(y+1)=0$

$y=0$ or $y=-1$

3. Replace the values of $y$ in equation (3) and solve for $x$

- For $y=0$

$x^2=6+y$

$x^2=6$

$x=\sqrt{6}$ or $x=-\sqrt{6}$

- For $y=-1$

$x^2=6+y$

$x^2=6-1$

$x^2=5$

$x=\sqrt{5}$ or $x=-\sqrt{5}$

So, the solutions of our system of equation are:

$x=\sqrt{6}, y=0\\ x=-\sqrt{6} , y=0\\x=\sqrt{5} , y=-1\\x=-\sqrt{5} , y=-1$

6 0

3 years ago

Read 2 more answers

The point nearest to the origin on a line is at (4, -4). Find the standard form of the equation of the line.

Scrat [10]

Just did a specific one of these; let's do the general case.

The point nearest the origin is (a,b).

The line from the origin through the point is

$bx - ay = 0$

The line we seek is perpendicular to this one. We swap the coefficients on x and y, negating one, to get the perpendicular family of lines. We set the constant by plugging in the point (a,b):

$ax + by = a^2 + b^2$