Question

Can someone help me learn Graphing a Parabola with a Vertex at the Origin?

Answer 1

My online class isn't explaining it well and I can't find videos with the problems I'm faced with.

Answer 2

These are one of the examples

Answer 3

Phantom Lord got this.

Answer 4

first, covert it into either \(Ax^2 + bx + c = 0 \) or \(f(x) = y = a(x-h)^2 +k \) where \(h, k \) is your vertex and \(a\neq 0\)
the first option is easier, the second option requires you to know how to complete the square

Answer 5

you can solve the vertex in the first one by using \(\large (x=\frac{-b}{2a},y = f(x)) \)

Answer 6

@nincompoop how would you put y=1/12x^2 in that first formula???

Answer 7

so if you have a standard form of \(y = 2x^2 - 4x + 3\) your vertex can be solved by identifying your coefficients
a = 2
b = 4
c = 3

vertex \(\large \frac{-b}{2a} = \frac{-(2)}{2(4)} = -\frac{1}{4}\)

then plug \(- \frac{1}{4} \) into all of the x in \(y = 2x^2+4x+3 \)

Answer 8

@nincompoop i think i'll just keep trying with youtube because you're confusing me even more

Answer 9

analyze the standard form \(y = Ax^2 + Bx +c \)
you have three terms
first term \(Ax^2 \)
second term \(Bx \)
third term \(c \)

Answer 10

ask yourself if the given equation has second or third term? if they don't show up it means it is zero anything +0 will remain the same

Answer 11

http://www.mathsisfun.com/geometry/parabola.html

Answer 12

http://mathbitsnotebook.com/Algebra1/Quadratics/QDVertexForm.html