Modeling With Quadratic Functions

[vaironxxrd]

Hello forum, Vaironl here.

I was on a trip for about 3 weeks and I missed many days of school, therefore I have many Algebra 2 questions so expect to see more annoying questions coming up since I have a midterm.

I just wanted to check if I'm doing this process right.

Given Vertex (2,-3) & a point (4,1)
Formula to use y = a(x-h)^2 + k

Solving:

1 = a(4-2)^2-3
1+3 = 4a
4 = 4a
a = 1

y = 1(x-2)^2 -3

 

[Deleted member 4993]

Hello forum, Vaironl here.

I was on a trip for about 3 weeks and I missed many days of school, therefore I have many Algebra 2 questions so expect to see more annoying questions coming up since I have a midterm.

I just wanted to check if I'm doing this process right.

Given Vertex (2,-3) & a point (4,1)
Formula to use y = a(x-h)^2 + k

Solving:

1 = a(4-2)^2-3
1+3 = 4a
4 = 4a
a = 1

y = 1(x-2)^2 -3

What was the question .... what were you supposed to find?

 

[burakaltr]

Hello forum, Vaironl here.

I was on a trip for about 3 weeks and I missed many days of school, therefore I have many Algebra 2 questions so expect to see more annoying questions coming up since I have a midterm.

I just wanted to check if I'm doing this process right.

Given Vertex (2,-3) & a point (4,1)
Formula to use y = a(x-h)^2 + k

Solving:

1 = a(4-2)^2-3
1+3 = 4a
4 = 4a
a = 1

y = 1(x-2)^2 -3

That's a Parabola.

y=a(x-h)**2+k
1=a(4-2)**2-3
1=a 2**2 - 3

4 a=4

a=1