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

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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) About Me  (Show Source):
You can put this solution on YOUR website!
The general form is +f%28x%29+=++a%2Ax%5E2+%2B+b%2Ax+%2B+c+
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 )
+-3+=+a%2A4%5E2+%2B+b%2A4+%2B+c+
(1) +-3+=+16a+%2B+4b+%2B+c+
------------------------------
For the vertex:
+-b%2F%282a%29+=+4+
(2) +b+=+-8a+
------------------------
( 6, -1 )
+-1+=+a%2A6%5E2+%2B+b%2A6+%2B+c+
(3) +-1+=+36a+%2B+6b+%2B+c+
-------------------------------
You have 3 equation with 3 unknowns, so
it's solvable
-------------------------------
Plug (2) into (1) and (3)
(1) +16a+%2B+4%2A%28+-8a+%29+%2B+c+%2B+3+=+0+
(1) +16a+-+32a+%2B+c+%2B+3+=+0+
(1) +-16a+%2B+c+%2B+3+=+0+
--------------------------------
(3) +36a+%2B+6%2A%28+-8a+%29+%2B+c+%2B+1+=+0+
(3) +-12a+%2B+c+%2B+1+=+0+
---------------------------------------
Subtract (1) from (3)
(3) +-12a+%2B+c+%2B+1+=+0+
(1) +16a+-+c+-+3+=+0+
------------------------------
+4a+-+2+=+0+
+4a+=+2+
+a+=+1%2F2+
and
(2) +b+=+-8a+
(2) +b+=+-4+
and
(3) +-12a+%2B+c+%2B+1+=+0+
(3) +-12%2A%28+1%2F2%29+%2B+c+%2B+1+=+0+
(3) +-6+%2B+c+%2B+1+=+0+
(3) +c+=+5+
----------------------------------
The equation is:
+y+=+%281%2F2%29%2Ax%5E2+-+4x+%2B+5+
----------------------------------
check:
( 4, -3 )
+-3+=+%281%2F2%29%2A4%5E2+-+4%2A4+%2B+5+
+-3+=+8+-+16+%2B+5+
+-3+=+-3+
OK
and
( 6, -1 )
+-1+=+%281%2F2%29%2A6%5E2+-4%2A6+%2B+5+
+-1+=+18+-+24+%2B+5+
+-1+=+-1+
OK
Here's the plot:
+graph%28+400%2C+400%2C+-10%2C+10%2C+-10%2C+10%2C+%281%2F2%29%2Ax%5E2+-+4x+%2B+5+%29+
Looks OK

Answer by MathTherapy(10552) About Me  (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: matrix%281%2C3%2C+y%2C+%22=%22%2C+a%28x+-+h%29%5E2+%2B+k%29

                          matrix%281%2C3%2C+-+1%2C+%22=%22%2C+a%286+-+4%29%5E2+%2B+-+3%29 ------- Substituting (4, - 3) for (h, k), and (6, - 1) for (x, y)

                          - 1 = 4a - 3

                          - 1 + 3 = 4a

                                2 = 4a

                               matrix%281%2C7%2C+2%2F4%2C+%22=%22%2C+%22a%2C%22%2C+or%2C+1%2F2%2C+%22=%22%2C+a%29

matrix%281%2C3%2C+y%2C+%22=%22%2C+a%28x+-+h%29%5E2+%2B+k%29

highlight_green%28matrix%281%2C3%2C+y%2C+%22=%22%2C+%281%2F2%29%28x+-+4%29%5E2+-+3%29%29 ----- Substituting 1%2F2 for a, and (4, - 3) for (h, k)

That's it!! Nothing complex or time-consuming, @ ALL!

Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


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

y+=+a%28x-h%29%5E2%2Bk

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

y+=+a%28x-4%29%5E2-3

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

y+=+%281%2F2%29%28x-4%29%5E2-3