Determine quadratic function given vertex and a point: Vertex: (1,1) and point: (2,4) Please explain Thanks!
Go back to the form: \[y=ax ^{2}+bx+c\] Now, the x-coord of the vertex = -b/2a, so \[1=\frac{ -b }{ 2a }\] Then: \[\frac{ 1 }{ 1 }=\frac{ -b }{ 2a }\] cross multiply and we get b = -2a
Now we can replace b with 2a: \[y=ax ^{2}+2ax+c\] Plug in the first point replacing x with 1 and y with 1 and you get an equation with just 2 unknowns: a and c Plug in the second point replacing the x and y again and you get a second equation with just 2 unknowns: a and c Now you can solve this system of equations as you have 2 unknowns and 2 equations.
Do you know how to solve a system of equations?
no
sorry, i didn't get a message when you answered Did you get: 3a + c = 1 8a + c = 4 ?
To solve a system, we want to be able to add 2 equations and have one of the variables "disappear". We could multiply that first equation by (-1). We get: -3a - c = -1 8a + c = 4 Notice that the signs in the first equation all changed because we mult by -1 Now add the equations downward. The c's should disappear.
thank you
I made an error.\[-\frac{ b }{ 2a } = 1\] So b = -2a ... sorry
ok
\[y=ax ^{2}-2ax+c\]
when we plug in (1,1), we get: -a + c = 1 When we plug in (2, 4), we get: 4 = 4a - 4a + c So: 4 = c Now you can use that to find a since: -a + c = 1 and c = 4, -a + 4 = 1 so a = ?
also, b = -2a, and now that we know 'a' we can use it to find 'b'.
would it be -2a?
-a + 4 = 1 a=3
so now what did you get for 'a' and 'b'?
sorry, 'c' and 'b'
a=3,b=-2a
c=4, b=2a?
b = -2a = -2(3) = _?
-6
remember I made that error and b = - 2a, not 2a
Good then you go back to \[y=ax ^{2}+bx+c \] and put 3 in for 'a', -6 in for b and 4 in for c. That gives you this specific quadratic equation A lot of work on that one!
I know thank you so much I need to work on my math skills more
if you plug in the given points to your resulting equation, you will find they both work. If you find the x-coordinate of the vertex, you will find it is 1
good luck I am a math teacher / tutor at a community college but i have developed nerve issues so I had to cut back and this is fun for me
What college is that? I should attend
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