Question

Which equation has a graph that is a parabola with a vertex at (–1, –1)?

Answer 1

Answer:

Equations of parabola are

1. $(y+1)^2=4a(x+1)$

2. $(y+1)^2=\:-4a(x+1)$

3. $(x+1)^2=4a(y+1)$

4. $(x+1)^2=\:-4a(y+1)$

Step-by-step explanation:

Given: Vertex of a parabola is ( -1, -1 )

To find: Equation of parabola

We Know, the Standard equation of all Parabola is given by

1. $(y-k)^2=4a(x-h)$ :- parabola open in positive x direction

2. $(y-k)^2=\:-4a(x-h)$ :- parabola open in negative x direction

3. $(x-k)^2=4a(y-h)$ :- parabola open in positive y direction

4. $(x-k)^2=\:-4a(y-h)$ :- parabola open in negative y direction

Where, ( h, k ) is coordinate of vertex and

             ( a, 0 ) for 1st eqaution , ( -a, 0 ) for 2nd equation, ( 0, a ) for 3rd equation & ( 0, -a ) for 4th  equation are coordinates of focus of parabola.

We are given with vertex's coordinates.

⇒ Focus remain same

Therefore, Equations of parabola are

1. $(y+1)^2=4a(x+1)$

2. $(y+1)^2=\:-4a(x+1)$

3. $(x+1)^2=4a(y+1)$

4. $(x+1)^2=\:-4a(y+1)$

Answer 2

"Answer: D. $y = (x + 1)^2 - 1$

Solution:

Concept:

The equation of the vertical parabola is of the form,

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

where (h, k) are the vertices of the parabola that is generated.

Using the same concept, here the correct answer is option D.

As only in this function, y = (x + 1)2 – 1, the vertices are (-1, -1). "