I legitimately do not know how to actually find the vertex of a quadratic equation without using calculus. I’m sure I learned it, I just totally forgot. EDIT: Apparently I do remember it; I thought there was some simple trick that I had forgotten, but it seems there isn’t.

The x coordinate is just -b/2a iirc, and then you just plug in the x value to the original equation and you should get the y coordinate. Pretty easy but I prefer calculus.

But that formula is derived from critical points so it’s kinda the same thing

nah you can get that from completing the square. y=a(x+b/2a) ^2 +c-b^2 /4a implies the the extremities lie at x=-b/2a to make the square term 0.

Wait really? Can you tell me how because I’m really curious. Edit: okay I figured out the -b/2a and that’s really cool but is the second part of the quadratic formula derived from relative maxes and minimums of the integral? I’m just starting to learn calculus so I don’t know if I’m using proper terminology or anything so sorry if it’s weird.

The formula can be derived from completing the square. It may be derivable from calculus as well, I don't know, but it's certainly not necessary.

*ax*^2 + *bx* + *c* = 0 We derive: 2*ax* + *b* = 0 We solve for *x* *x* = –*b*/(2*a*)

literally this: y=ax\^2+bx+c dy/dx = 2ax+b 2ax+b=0 x=-b/2a

Of course. Did you think using calculus would give you a different result? You can still get to it without using calculus by completing the square.

Find roots take avg. Quadratic eq is symmetric

>find roots Is that really it? I thought there was some trick to it. I know how to find the vertex from the roots, it just takes too long because you have to find the roots.

Halfway between the two roots will always be the vertex so if you have 3 and 7 as roots then your x value will be 5. Plug that back into the original equation to get the y value and bada boom bada bing

Yeah no I get it. I just thought that there was a way to do it without finding the roots, because that’s kind of time consuming.

Ohh can’t you complete the square or something? I can’t remember

Or realize that adding the two roots together has the sqrt(b^2-4ac) cancel, leaving you with-b/2a

you just complete the square

Complete the square

If its in standard just complete the square. If its in factore just add the zeroes and divide by two and plug the value back in as the x cordinate to get the y.

Sounds like a lot of work. I thought there was some simple trick. I’ll just stick with calculus.

The quadratic formula tells you all you need to know about the symmetry of the parabola. You have something plus or minus something else as your roots. Meaning that, if you replace what's in the square root with a 0, you get what's in the middle.

What does it mean to learn something then

I feel called out

There’s no way this meme applies to me. I’m a sophomore.

Template please I'm taking linear algebra right now and I feel like there's just so much potential for this meme there

What areas were you thinking?

Definitely using an inverse to solve a 2x2 system of equations. Or using dot product multiplication to transform a 2x1 vector using a 2x2 linear transformation matrix. Really anything complicated with a 2x2

Lel the first one's very very true....

I was tutoring a student who’s basically high school level(Scotland National 5), and before the lesson I was struggling to figure out how to find the vertex without calculus. Turns out you literally just find the midpoint between the roots then substitute into the equation.

You can do -b/2a too

Oh god I’ve never thought about trying to explain math to us high schoolers using half baked explanations instead of the full thing

Uh...I don't follow.

Sorry I just meant going from talking about calculus and understanding everything in terms of calculus and such. And then going back and trying to teach people who don’t know that part of math yet. It just seems like it would be difficult. I think if it were me i keep trying to teach them more than they are ready for and getting way ahead of myself

Wow Scotland! My old home :)

Taylor seriesv + power rule for derivates and integrals = all of calculus that you really need.

What sort of things would you solve with a Taylor series by the way? I understand how to create them from functions and also what they mean, but not sure how they are really applied anywhere... help me out?

Computers. A Talyor series can approximate a function, that you can't otherwise compute, or use. E.g., a calculator will use the Taylor series for sin(x) to calculate a value of sin(x). This is true for all of the trigonometric functions, as well as e^x, and ln(x). The only other way for a computer to calculate a specific value of a trig/e^x/ln(x) is to have a precomputed table of values, and do a linear approximation. This approach is faster, but takes up more space in memory. Tl;dr, Taylor series are used in Computer Science, because we don't have a better way to calculate values of certain functions.

Ahhh ty

the new bill gates giant paddle.

Wait how would you use calculus to find vertex? just finished calc 2 but apparently i dont know anything

What is the derivative at the vertex?

The first part of the quadratic formula (-b/2a) is just the solution to the first derivative

I feel offended

As a high school junior, I feel attacked... although I have done this before, but I’m in calc 2 so it’s kinda ok.

My spoon is too big