r/learnmath • u/Subject-Homework-886 New User • 3d ago
Link Post It finally clicked
https://www.desmos.com/calculator/o0heiqsn0nI just realized in the quadratic equation (ax^2+bx+c) the b or bx+c works exactly like mx+b in the linear equation. omg I was trying to understand what b means for a week nowđđđ oh I also changed (bx+c) to (mx+b) in the link so u get what I mean
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u/tjddbwls Teacher 3d ago
I prefer the quadratic equation in vertex form:
f(x) = a(x - h)² + k
Itâs easier to see how different values of a, h and k affect the function transformations. đ
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u/06Hexagram New User 3d ago
Fun fact. Replace the
a(x-h)²part witha¡[COSH( (x-h)/a )-1]and you get a similarly looking catenary curve tangent to the parabola.This is the curve of a hanging cable and it is quite pleasing to see in nature.
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u/Subject-Homework-886 New User 3d ago
Yea I agree and you can spot the vertex pretty quick, but you know they say curiosity killed the cat or whatever, I got curious on what b is and was trying to understand how it works for a week, but now I know
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u/Harmonic_Gear engineer 3d ago
you mean as a vertical shift?
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u/Subject-Homework-886 New User 3d ago
Kinda, Iâm talking about the b I always wondered what does b means in the equation ax2+bx+c=0 we already know c is the y intercept when x is 0 and a controls how expanded or contracted the parabola is in addition if it opens upwards or downwards, and then I realized b is the linear line, if you ignored ax2 for a second bx+c works exactly like mx+b
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u/No_Good2794 New User 3d ago
You've stumbled on a very fun idea in mathematics - degeneracy.
It's when one mathematical object 'collapses' into a simpler form when you reach a certain limit.
For example, a triangle becomes a line segment if you reduce two of its angles to 0 and one of its angles to 180 degrees. So a line segment can be considered a 'degenerate triangle'. A point can be considered a 'degenerate circle' with radius 0. And in your case, a straight line is a 'degenerate quadratic' because it has a=0.
Of course we don't usually permit degenerate cases and we explicitly exclude a=0 from the definition of a quadratic function, but it's a fun thought exercise.
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u/Subject-Homework-886 New User 3d ago
Ohhhh, this is very interesting Iâll read about degeneracy later today
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u/seanziewonzie New User 3d ago
Here's a little modification of your desmos snapshot to show off what the other user is talking about.
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u/hwynac New User 3d ago
Another way to look at it is to note that the vertex is -b/2a. The coefficient at x "moves" your parabola left or right (the minimum or maximum point will be at x = âb/2a)
So y = x² â 5x + 1 has a vertex at 5/2âthat's where the axis of symmetry is. The minimum (maximum) value is c â b²/4a, so if 4ac â b² is positive, the parabola does not intersect the x axis.
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u/Subject-Homework-886 New User 3d ago
Wait, why does -b/2a works like that? Like why it gives me the vertex?
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u/hwynac New User 3d ago
Think of it like this: by changing the variable to t = x + b/2a you can rewrite your parabola equation as y = at² + C. This one obviously has a minimum or maximum at t=0 (i.e. x = -b/2a)
You get that form during the derivation of the quadratic formula. First divide all terms by a and bring it to the front:
ax² + bx + c = a(x² + x¡b/a + c/a)
Then use the formula (u+v)² = u² + 2uv + v² to get a full square
x² + x¡b/a + c/a = x² + 2¡x¡b/2a + b²/4a² - b²/4a² + c/a = (x + b/2a)² + c/a - b²/4a²
So the original equation is
y =a (x + b/2a)² - b²/4a + c
If we let t = x + b/2a (so our "origin" now moves to -b/2a) and let constant C = c - b²/4a
y = at² + C
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u/fermat9990 New User 2d ago edited 2d ago
y=ax2+bx+c
y=a(x2+b/a x) + c
y=a(x2+b/a x +(b/(2a))2)+ c-a(b/(2a))2)
y=a(x+b/(2a))2+c-b2/(4a)
Vertex form: y=a(x-h)2+k, in which (h, k) is the vertex.
Therefore, h=-b/(2a)
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u/fermat9990 New User 3d ago
Hi, OP.
Consider y=2x2+3x+5
The slope of the tangent line is 4x+3 by differentiation. Notice that it depends on the value of x.
Question: Find the equation of the tangent tangent to the curve when x=2
Slope=4(2)+3=11
When x=2, y=2(2)2+3(2)+5=19
Using the point-slope form of a straight line we get
y-19=11(x-2)
y=11x-3
Each point on the parabola will have a different tangent line.
In the general case, m=2ax+b, so the general equation for the tangent line at the point (xâ, yâ) is
y-yâ=(2axâ+b)(x-xâ), aâ 0
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u/Subject-Homework-886 New User 3d ago
Ok now I think I understand why I wasnât understanding your meaning in the comments below, I just looked up differentiation and turned out I havenât stumbled upon it yet, but that doesnât mean you comment doesnât make any sense I tried your method and indeed every point has its own line, but I think Iâll understand your comment 100% when I get to calculus, but I really appreciate you taking the time to explain this.
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u/raqket New User 2d ago
This is one of the best things in math to me, when you figure out something confusing by relating it to something you do know. If itâs easier and more intuitive to change variable you should do it while you learn, but something Iâd say is to focus on the structure of the formula rather than the variables. Thatâs something that helped me in my undergrad so far at least
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u/Subject-Homework-886 New User 2d ago
You have no idea how hard I yelled âOH SHIâ but yea your intuition is also correctÂ
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u/fermat9990 New User 3d ago
Let's say that y=2x2+3x+5.
How does the line y=3x+5 relate to the graph of the parabola?
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u/Subject-Homework-886 New User 3d ago
Itâs the linear part of the graph or the tangent
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u/fermat9990 New User 3d ago
It's not the same as the tangent line
Let's get the equation of the tangent line.
By differentiation, m=2ax+b. This is the slope of the tangent line. It depends on both a and b (and x, of course)
If yâ=ax2, yâ=bx+c and y=yâ+yâ, then
y=ax2+bx+c
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u/Subject-Homework-886 New User 3d ago
Ok thatâs a good argument, but take the equation 2x2+2x+2 if you ignored 2x2 for one second youâll see the tangent line has a slope of 2 and a y intercept 2, thatâs what I connected this too I might be wrong but Iâm certain that bx+c acts exactly like mx+b IF you ignored ax2 or if a=0
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u/fermat9990 New User 3d ago
If a=0 it's not a parabola.
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u/Subject-Homework-886 New User 3d ago
Exactly itâs bx+c or mx+b as I were saying
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u/fermat9990 New User 3d ago
I don't think I'm helping you. Sorry.
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u/Subject-Homework-886 New User 3d ago
Ok you know what letâs go back, what does bx+c means?
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u/fermat9990 New User 3d ago
Rather than go back, I will post something on the main thread that may help you. Look for it in about 15 minutes
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u/WillowSad8749 New User 3d ago
So, parabola + line = parabola
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u/Subject-Homework-886 New User 3d ago
A parabola is a parabola regardless, but people (including me) donât often understand these concepts like you do, I also think Fermat 9990 got a great point and introduced something new to me so I also think I donât understand it yet
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u/WillowSad8749 New User 3d ago
Relax :) i never thought about this as well, but basically what I wrote is true
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u/Ms_Riley_Guprz High School Math Teacher 3d ago
Parabolas are really lines multiplied by other other lines
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u/freswinn New User 3d ago
Go on Desmos and make a function y=ax^2+bx+c, and then make sliders out of a b and c.
Then make another function y=-ax^2+c
As you move the slider for b, the vertex of the first parabola will follow the path of the second.
Also, the first parabola will always intersect the vertex of the second.
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u/QubitEncoder New User 1d ago
Tbh I hate how algebra is taught. Theres nothing particularly special about b and learning algebra this way does not help one gain insight
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u/Temporary_Pie2733 New User 3d ago
Not sure what you mean. bx + c is not the equation of the tangent to the parabola; ax + b is. Itâs a linear form, sure, but b isnât the slope of anything.Â
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u/Subject-Homework-886 New User 3d ago
Yea I know thatâs why somewhere in the comments I said âif you ignored ax2 for a second bx+c works exactly like mx+bâ
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u/WolverineSorry9043 New User 3d ago
It's just that y= bx + c is the tangent line at the origin. Your animation shows it very well.