How to find the Equation of a Parabola using its Vertex and its y-intercept - Tutorial 1

We learn how to find the equation of a parabola using the coordinates of its vertex (maximum, or minimum, point) as well as its y intercept.
Given a parabola's vertex, with coordinates (h,k), we can write the standard equation y=ax^2 + bx + c can be re-written in the form : y = a(x - h)^2 + k
We see how this can be used to find a parabola's equation.

[@avery9879]

Just started precalc for sophomore year and I needed a refresher, thanks!

[@RadfordMathematics]

Thanks for the comment! Really glad it helped 😊

[@ricardo-c4l7l]

Thank you for this. Good teaching, like a fine wine, improves with time.

[@tristaniga6689]

Could you explain how you expanded the vertex form in the example? I don't know how you got "(x²-4x+4)". 🙏

[@MathewsDozola-ce2hv]

You're blessed in explanation. Classic 🔥🔥🔥🔥

[@anaghaharinair1562]

Thank you so much

[@RadfordMathematics]

Really glad it helped 😊

[@chimichangaking5640]

I’m trying to find the equation of a parabola but this doesn’t work because the vertex is (0,0) what else can i do?

[@RadfordMathematics]

Thanks for your comment 😊
If a parabola passes through (0,0) then it’s equation has to look like : y=ax^2 your only job is to then find the value of a and for the you need the coordinates of one other point it passes through.
So, for example say you have a parabola that passes through (0,0) and it also passes through (1,2) then since it has an equation y=ax^2 you know that when x=1 y must equal to 2. So you can write:
2=a.1^2
2=a.1
2=a
So a=2 and the parabola’s equation would be:
Y=2x^2
Hope that helps 😊

[@derekweiland1857]

How can I find the equation of a parabola if I only have the axis of symmetry and the y-intercept?

[@Kappa1t]

im confused where you got the numbers to plug in at 4:20

[@KeenanWilliams777]

(x-1)^2 = (x-1)(x-1).
to multiply (x-1) and (x-1), distribute the terms from the left side and multiply them to the right side
= (x * x) + (x * -1) + (-1 * x) + (-1 * -1)
=(x^2-1x-1x+1)
=(x^2-2x+1)

[@beetlebro4312]

Can you teach me how to find the equation with the Vertex, and both x and y intecept?

[@RadfordMathematics]

Hi there!
Thanks for your comment and question 😊.
To find the equation using the vertex : that’s what you saw in this video.
Using the x intercepts in the other hand you should use root factoring 🤓.
I show how to do that here : ruclips.net/video/pxrZkMHLVT4/видео.html
Hope that helps!
Take good care 🌱

[@beetlebro4312]

@@RadfordMathematics hello sir I actually got the given wrong, it was actually a point that passes through the parabola, not the x and y intercepts. So I solved it. Thank you for replying to me though 🙏

[@sk7192]

cheers

[@JeremiahKaro-en9rm]

My question is
Why you used x-1 instead of x+1

[@yashvim2319]

vertex form is written as y = a (x - h)^2 +k, where (h,k) is the vertex. In the equation though, it is written as -h, so you must change the value when putting it into the equation. Hope this helps