SOLUTION: How do i make a quadratic equation from vertex point (4,-3) and point (6,-1)?

Question 1148201: How do i make a quadratic equation from vertex point (4,-3) and point (6,-1)?
Found 3 solutions by josmiceli, MathTherapy, greenestamps:
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!
The general form is
The y-component of ( 6, -1 ) is greater than the
y-component of the vertex, so I know the vertex is
a minimum ( just an observation )
----------------------------------------------------
( 4, -3 )

(1)
------------------------------
For the vertex:

(2)
------------------------
( 6, -1 )

(3)
-------------------------------
You have 3 equation with 3 unknowns, so
it's solvable
-------------------------------
Plug (2) into (1) and (3)
(1)
(1)
(1)
--------------------------------
(3)
(3)
---------------------------------------
Subtract (1) from (3)
(3)
(1)
------------------------------

and
(2)
(2)
and
(3)
(3)
(3)
(3)
----------------------------------
The equation is:

----------------------------------
check:
( 4, -3 )

OK
and
( 6, -1 )

OK
Here's the plot:

Looks OK

Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!
How do i make a quadratic equation from vertex point (4,-3) and point (6,-1)?

Vertex form of a quadratic:
------- Substituting (4, - 3) for (h, k), and (6, - 1) for (x, y)
- 1 = 4a - 3
- 1 + 3 = 4a
2 = 4a

----- Substituting for a, and (4, - 3) for (h, k)
That's it!! Nothing complex or time-consuming, @ ALL!
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

Start with the vertex form of the equation, as tutor @MathTherapy did:

With the vertex given as (4,-3), the equation in this form is

Then here is an alternative method for determining the value of a to complete the equation.

(1) The point other than the vertex that is given is 2 units to the right of the vertex.
(2) If a were 1, so that the graph behaved like y=x^2, then the y value 2 to the right of the vertex would be 2^2=4 greater than the y value at the vertex.
(3) But the y value 2 to the right of the vertex is only 2 greater than the y value at the vertex.

Since the y value increased only half as much as it would have increased if a were 1, a must be 1/2.

So the equation is

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