## briana.img one year ago Can someone help me learn Graphing a Parabola with a Vertex at the Origin?

1. briana.img

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

2. briana.img

These are one of the examples

3. Mr_Perfection_xD

Phantom Lord got this.

4. briana.img

5. nincompoop

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

6. nincompoop

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

7. briana.img

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

8. nincompoop

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 $$y = 2x^2+4x+3$$

9. briana.img

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

10. nincompoop

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

11. nincompoop

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

12. nincompoop
13. nincompoop