Find two equations of parabolas whose vertex is ( -1,2) and passing through (0,5)

Question

im having trouble trying to find an answer for this question if someone could help me out i would be extremely grateful thank you

Answer 1

Hi Nicholas!

Since the vertex was given I would first put it in vertex form. Then using the point you will be able to find the value a.

y= a(x-h)² + k V(-1,2) Point ( 0,5)

y= a(x+1)² + 2

Now replace (0,5) with y and x

5= a (0+1)² + 2

Solve for a

5 = a(1) + 2

5 =a +2

3= a

now you have it in vertex form: y= 3(x+1)² + 2

I hope this helps you. Have a great day!

Mrs. S

Answer 2

The easiest way to start this problem would be to start with the general form of a parabola with a vertex at (-1, 2).

The vertex form of an equation for a parabola that opens parallel to the y-axis (i.e. one that looks like a U or ∩) with a vertex of (x₁, y₁) is

y - y₁ = a(x - x₁)²

where a is a constant factor (i.e. just a number such as 1, 1.4, 2, etc.)

Plugging our vertex in gives

y - 2 = a(x - (-1))²

We can plug in our values of the point it passes through (0, 5) to find the value of a

5 - 2 = a(0 + 1)²

Which gives us

3 = a

So the first equation would be

y - 2 = 3(x + 1)²

Now, it is impossible to have another parabola that opens upwards or downwards (U/∩), because a parabola is symmetric (with the vertex lying on the axis of symmetry). This means that point (-2, 5) would also be on a parabola opening up/downwards, and three points define a parabola.

Then we need to look for a parabola that opens to the left or right (⊃ or ⊂).

Then the x and y of the initial form of the equation would be switched:

x - x₁ = a(y - y₁)²

Plugging in values again yields

x - (-1) = a(y - 2)²

And once again, we solve for a using the point (0, 5) on the parabola

0 + 1 = a(5 - 2)²

1 = 9a

a = 1/9

So our second equation is

x + 1 = 1/9 (y - 2)²

or

9(x + 1) = (y - 2)²