Question

Derivatives of quadratic equations :)

Answer 1

\[f(x)=ax^2+bx+c \\ \text{ so you want to differentiate } f\]
First question: Are you doing short cuts or the definition way?

Answer 2

I'm not sure, I just would like to learn how to do this so I'm not always confused lol

Answer 3

You are probably doing definition I bet.

Answer 4

\[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \text{ or } f'(x)=\lim_{z \rightarrow x} \frac{f(z)-f(x)}{z-x}\]
you can use either one of these
these are basically the same thing
they are the same thing
it can be shown with a substitution
Let's play with the first one...

Answer 5

\[f(x)=ax^2+bx+c \\ f(x+h)=a(x+h)^2+b(x+h)+c\]
To find f(x+h) notice I just replace all the x's in the first expression with (x+h)'s.

Answer 6

the numerator in my definition is f(x+h)-f(x)
so I could go ahead and look at simplifying that
first I will state what is without the simplification :
\[f(x+h)-f(x)=[a(x+h)^2+b(x+h)+c]-[ax^2+bx+c]\]
The next part is to expand and cancel.

Answer 7

Can you expand (x+h)^2

Answer 8

\[x^2+2hx+h^2\]

Answer 9

cool so I'm going to put that in place of (x+h)^2

...
\[f(x+h)-f(x)=[a(x^2+2xh+h^2)+b(x+h)+c]-[ax^2+bx+c] \\ \text{ now distributing a bit } \\ f(x+h)-f(x)=ax^2+2axh+ah^2+bx+bh+c-ax^2-bx-c\]
do you have a problem with my distributing step?

Answer 10

Nope, easy peasy

Answer 11

you should see some things that cancel (or zero out)
like ax^2-ax^2=0
and bx-bx=0
and c-c=0

Answer 12

\[f(x+h)-f(x)=\cancel{ax^2}+2axh+ah^2+\cancel{bx}+bh+\cancel{c}-\cancel{ax^2}-\cancel{bx}-\cancel{c}\]

Answer 13

So we left with \[f(x+h)-f(x)=2axh+ah^2+bh\]

Answer 14

right!

Answer 15

now let's go back to our definition
this was only the numerator

Answer 16

Okay :)

Answer 17

\[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{2axh+ah^2+bh}{h}\]

Answer 18

so... notice that every term on top has a common factor of h
and the denominator also has a common factor h

Answer 19

\[f'(x)=\lim_{h \rightarrow 0} \frac{h(2ax+ah+b)}{h(1)}\]

Answer 20

Oh!

Answer 21

we can write as:
\[f'(x)=\lim_{h \rightarrow 0} \frac{h}{h} \cdot \frac{2ax+ah+b}{1}\]

Answer 22

which the h's cancel, right?

Answer 23

yep and that whole h on the bottom was the thing that was giving us problems
but guess what?

Answer 24

it is no longer there because we canceled that mean h guy on bottom

Answer 25

:)

Answer 26

so we can plug in 0 now

Answer 27

for h that is

Answer 28

and it leaves us with 2ax+b?

Answer 29

\[f'(x) =\lim_{h \rightarrow 0} (2ax+ah+b)=2ax+a(0)+b=2ax+0+b=2ax+b\]
you are right

Answer 30

Is that it?

Answer 31

\[f(x)=ax^2+bx+c \text{ gives us } f'(x)=2ax+b\]

Answer 32

yes now you can find the derivative of any quadratic

Answer 33

you actually have a formula for it now
we just found a formula for it above

Answer 34

Awesome! Didn't know it was that simple...the video I watched on it made it seem so confusing lol

Answer 35

So there is never a c...

Answer 36

\[f(x)=ax^2+bx+c \text{ gives } f'(x)=2ax+b \\ \text{ example } \\ \text{ say the question is find } f' \text{ given } f(x)=5x^2+3x-4 \\ \text{ we only need to know } a \text{ and } b \\ f'(x)=2(5)x+3=10x+3\]

Answer 37

Well the derivative of a constant will always be zero.

Answer 38

Schnazzy!

Answer 39

Because the slope of y=a constant is zero.

Answer 40

Thank you so very much! :)

Answer 41

Np.
So this is exactly what you wanted right?
Do you have anymore questions about differentiating a quadratic?

Answer 42

Yes and no, I don't! :)

Answer 43

Wait...perhaps one question...this works for all quadratic equations?

Answer 44

Alright. I might do one more thing before I leave... A different strategy...
Instead of expanding...grouping and factoring...
\[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{ a(x+h)^2+b(x+h)+c-(ax^2+bx+c)}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{a[(x+h)^2-x^2]+b((x+h)-x]+[c-c]}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{a[(x+h)^2-x^2]+b[(x+h)-x]+0}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{a[(x+h-x)(x+h+x)+b[x+h-x]}{h} \\ \\ f'(x)=\lim_{h \rightarrow 0} \frac{ah(2x+h)+bh }{h} \\ f'(x)=\lim_{h \rightarrow 0} a(2x+h)+b \\ f'(x)=a(2x+0)+b =2ax+b\]

Answer 45

and yes
to your question
you do know a quadratic function is in this form right:
\[f(x)=ax^2+bx+c ?\]

Answer 46

yes!

Answer 47

and if you do then yes to your question

Answer 48

I love quadratic equations....and finding the solutions using the quadratic formula too :)

Answer 49

do you know that you can find the x-coordinate of vertex of the parabola using the derivative

Answer 50

I did not!

Answer 51

Yep. You will probably learn this later. But You can use the derivative to either find the max or min of a function.
You can set the derivative equal to zero and find where it does not exist...These will be critical numbers...

So you have
\[f(x)=ax^2+bx+c \\ f'(x)=2ax+b \\ f'(x)=0 \text{ gives you the equation } 2ax+b=0 \\ \\ \text{ solving for } x \text{ gives you } x=\frac{-b}{2a}\]

Answer 52

This is exactly what we would get if we did the whole completing the square thing.

Answer 53

\[f(x)=ax^2+bx+c \\ f(x)=ax^2+\frac{a}{a}bx+c \\ f(x)=a(x^2+\frac{1}{a}bx)+c \\ f(x)=a(x^2+\frac{b}{a}x)+c \\ f(x)=a(x^2+\frac{b}{a}c+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ f(x)=(x+\frac{b}{2a})^2+c-a(\frac{b}{2a})^2 \\ \text{ where the vertex is } (-\frac{b}{2a}, c-a(\frac{b}{2a})^2)\]

Answer 54

But anyways all of that is extra.

Answer 55

Thank you again :)

Answer 56

np