find the vertex
- Thread starter haliebre
- Start date
Hello, haliebre!
For the parabola, \(\displaystyle y\:=\:ax^2\,+\,bx\,+\,c\), the vertex is at: \(\displaystyle \:x\;=\;\frac{-b}{2a}\)
This parabola has: \(\displaystyle \,a\,=\,-1,\;\;b\,=\,6,\;\;c\,=\,-8\)
So we have: \(\displaystyle \,x\:=\:\frac{-6}{2(-1)}\:=\:3\)
Then: \(\displaystyle \,y\:=\:-3^2\,+\,6(3)\,-\,8\:=\:1\)
The vertex is: \(\displaystyle \,(3,\,1)\)
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[rant]
At this point, someone will counter with "I'd rather not have the student
\(\displaystyle \;\;\)blindly memorize a list of formulas. \(\displaystyle \;\)I prefer Understanding."
Fine, be that way . . .
But this is just one teensy formula in a future of hundreds of formulas.
When this student gets to Differential Equations and has: \(\displaystyle \,\frac{d^2y}{dx^2}\,-4\frac{dy}{dx}\,+\,5y\;=\;0\)
I hope he/she can find the roots: \(\displaystyle \,m\:=\:2\,\pm\,i\) without completing-the-square.
I hope he/she has blindly memorized: \(\displaystyle \,\frac{d}{dx}(\sin x) \,=\,\cos x\)
\(\displaystyle \;\;\)instead of going through \(\displaystyle \,\lim_{h\to0}\frac{f(x+h)\,-\,f(x)}{h}\) every time.
And I certainly hope that \(\displaystyle \,7\,\times\,8\:=\;56\) is already hard-wired in his/her brain.
Understanding is a good thing.
I teach all concepts with the goal of Understanding.
And when it's time to move on, I do so.
I will show them formula, shortcuts, theorems, etc. to facilitate the work.
\(\displaystyle \;\;\)(And this is not my idea . . . they are in the textbooks, mind you.)
This is not Blind Memorizing . . .it's called Learning Mathematics!
[/rant]
Didn't your teacher show you the "vertex formula"?
For the parabola, \(\displaystyle y\:=\:ax^2\,+\,bx\,+\,c\), the vertex is at: \(\displaystyle \:x\;=\;\frac{-b}{2a}\)
This parabola has: \(\displaystyle \,a\,=\,-1,\;\;b\,=\,6,\;\;c\,=\,-8\)
So we have: \(\displaystyle \,x\:=\:\frac{-6}{2(-1)}\:=\:3\)
Then: \(\displaystyle \,y\:=\:-3^2\,+\,6(3)\,-\,8\:=\:1\)
The vertex is: \(\displaystyle \,(3,\,1)\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
[rant]
At this point, someone will counter with "I'd rather not have the student
\(\displaystyle \;\;\)blindly memorize a list of formulas. \(\displaystyle \;\)I prefer Understanding."
Fine, be that way . . .
But this is just one teensy formula in a future of hundreds of formulas.
When this student gets to Differential Equations and has: \(\displaystyle \,\frac{d^2y}{dx^2}\,-4\frac{dy}{dx}\,+\,5y\;=\;0\)
I hope he/she can find the roots: \(\displaystyle \,m\:=\:2\,\pm\,i\) without completing-the-square.
I hope he/she has blindly memorized: \(\displaystyle \,\frac{d}{dx}(\sin x) \,=\,\cos x\)
\(\displaystyle \;\;\)instead of going through \(\displaystyle \,\lim_{h\to0}\frac{f(x+h)\,-\,f(x)}{h}\) every time.
And I certainly hope that \(\displaystyle \,7\,\times\,8\:=\;56\) is already hard-wired in his/her brain.
Understanding is a good thing.
I teach all concepts with the goal of Understanding.
And when it's time to move on, I do so.
I will show them formula, shortcuts, theorems, etc. to facilitate the work.
\(\displaystyle \;\;\)(And this is not my idea . . . they are in the textbooks, mind you.)
This is not Blind Memorizing . . .it's called Learning Mathematics!
[/rant]
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,990
I totally disagree with that.soroban said:
I call that “being hopelessly out of touch with what is going on in mathematics reform.”
Why in the world should any one learn a formula for the vertex of a parabola? Actually graphing utilities have made that obsolete.
I can see wanting a student to be able to apply completing a square.
What a basic principle to know how to use. I am all for back to basics!
But like partial fractions, anyone will have a very hard time convening me that knowing a formula for the vertex of a parabola is basic to any understanding.
On a trip to America in the 1930’s, Einstein was asked by a reporter to tell him the value of pi to ten decimal places. His answer “Good lord man I do not remember anything that I can look up”.
pka said:
By your reasoning, the formula for the circumference of circle must wait until Calculus 2 or 3,
\(\displaystyle \;\;\)when Arc Length is taught . . . that is, we can't use \(\displaystyle \,C\:=\:\pi d\) until we understand the basics.
If we do, we're hopelessly out of touch with what's going on in mathematical reform.
\(\displaystyle \;\;\)And heaven forbid, we ever use \(\displaystyle \,A\,=\,\frac{1}{2}bh\)
Why must we learn about Riemann sums when our calculators will integrate anything?
Why do we bother teaching long division when everyone uses calculators anyway?
Funny, this issue always arises with the "vertex of a parabola" formula.
\(\displaystyle \;\;\)What is it about this one formula that sets people off?
It's okay to memorize that: \(\displaystyle \,7\,\times\,8\:=\:56\,\) or that: \(\displaystyle A\:=\:\pi r^2\)
\(\displaystyle \;\;\)but somehow \(\displaystyle \,x\,=\,\frac{-b}{2a}\) is an assualt on human sensibilities.
It evidently is an totally useless piece of information and should be stricken from all textbooks.
\(\displaystyle \;\;\)Go figure . . .