Quadratic function: find function from two points on quadratic

[Zulgok]

Hello, I have 2 coordinates from quadratic function (2;-2), (0;-1). I have to find a, b and c. y=ax^2+bx+c - as a first equation I have now got -2=4a+2b+c. Second equation -1=c. Unfinished result is -1=4a+2b. What do I do next to find a and b ? a>0

 

[Ishuda]

Hello, I have 2 coordinates from quadratic function (2;-2), (0;-1). I have to find a, b and c. As a first equation I have now got -2=4a+2b+c. Second equation -1=c. Unfinished result is -1=4a+2b. What do I do next to find a and b ? a>0

If you only have two points, you can fit that with a linear equation with a slope s of
s = [-1 - (-2)] / [0 - 2] = -0.5
and resulting equation of
y(x) = -1 - 0.5 x.
You can make that a quadratic equation if you like by adding another piece, i.e.
y(x) = -1 - 0.5 x + a x (x-2) = a x² - 0.5 (4 a + 1) x - 1
where a is any value you chose other than zero [since that would make it a linear equation, not a quadratic equation.]

The way you are doing it looks like you want an equation of
y(x) = a x² + b x + c
which would give the equations you have:
(1) -2 = 4 a + 2 b + c
(2) c = -1
Plugging (2) back into (1) gives
-2 = 4a + 2 b -1
or
b = -0.5 (4a +1)
and we have
y(x) = a x² - 0.5 (4 a + 1) x - 1
as above.

 

[stapel]

I have 2 coordinates from quadratic function (2;-2), (0;-1). I have to find a, b and c. y=ax^2+bx+c

Unless you are provided other information, you cannot find "the" quadratic function through the two points, because there are more than one quadratic functions which will fit through the data.

Please reply with the full and exact text of the exercise, the complete instructions, a description of any other information (a picture, a table, etc). Thank you!

 

[Zulgok]

Here is a photo of it :

< link to objectionable page removed >

Asks me to write down function y=f(x).

 

[Ishuda]

Here is a photo of it : <link removed> . Asks me to write down function y=f(x).

The image gives one more bit of information which you hadn't given and that is that the vertex is at x=2. See
http://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html
for example to find out more about the vertex form of the quadratic. Given that the location for the vertex is (2,-2), we have
y(x) = d (x-2)² - 2.
thus y(2)=-2. To find the value d, plug in the the other point, i.e. (0,-1) which gives
-1 = d (0-2)² - 2.
So what is d. Given that, what are the values of a, b, and c for the form
y(x) = a x² + b x + c

 

[Zulgok]

Hm, we didnt cover this thing in class. Thank you for help, ill try learning vortex now.

 

[Ishuda]

Hm, we didnt cover this thing in class. Thank you for help, ill try learning vortex now.

You could also get the same answer by using a third point on the curve such as (4,-1)