How to Find the Vertex of a Quadratic Function

[@mk4630]

I'm doing my homework right now and I was about to look up the answer but this video popped up and gave me exactly what I needed lmao

[@hoshiparrot5834]

THIS MAKES SO MUCH MORE SENSE THAN HOW MY PROFESSOR DID IT THANK YOU

[@Larfage]

Great video! Simple and to the point. Fun fact: this formula comes from taking the derivative (calculus concept) of the equation - you then solve the derivative for 0 (because the slope of the tangent line will be 0 at the vertex). Solving this gives you the x-coordinate of the vertex, which you then plug in to the original equation to find the y-coordinate. Ta-da, you now have the vertex

[@AryanSingh-iu9vt]

You can avoid calculus by completing the square and noticing the fact that the square of any real number is non negative so you can equate the squared term =0 to obtain the min/max !

[@nostromoglasseye8340]

what about an equation such as x^2 - 1, or 4 - x^2? these have a vertex observable in the parabola graphed, but b=0

[@nostromoglasseye8340]

though it is in fact -1 and 4 respectively ...

[@twowombmates]

it would be helpful to show graphically what the vertex is of this parabola… M

[@dylanparker130]

I believe vertices are "extrema of curvature", if that helps?

[@dylanparker130]

By pure coincidence, I was reading a paper today which mentioned the vertex of a curve! It described any vertex as a "point of extreme curvature", which I guess means a curve's vertices are its points of highest and lowest curvature?

[@Stvrfiire_]

Is this true always?

[@TheMathSorcerer]

yup, super powerful, just make sure it's written like f(x) = ax^2 + bx + c, where a !=0 obviously:)

[@Stvrfiire_]

@@TheMathSorcererHaha okay thank you math sorcerer!

[@NINJA-tf6bf]

I'm the first one!

[@MGPablo]

just use the first derivative test for min/maxima, find x and plug back into f(x) 🎉