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haleyelizabeth2017
 one year ago
Derivatives of quadratic equations :)
haleyelizabeth2017
 one year ago
Derivatives of quadratic equations :)

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freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(x)=ax^2+bx+c \\ \text{ so you want to differentiate } f\] First question: Are you doing short cuts or the definition way?

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0I'm not sure, I just would like to learn how to do this so I'm not always confused lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.3You are probably doing definition I bet.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)f(x)}{h} \text{ or } f'(x)=\lim_{z \rightarrow x} \frac{f(z)f(x)}{zx}\] 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...

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(x)=ax^2+bx+c \\ f(x+h)=a(x+h)^2+b(x+h)+c\] To find f(x+h) notice I just replace all the x's in the first expression with (x+h)'s.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3the 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.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Can you expand (x+h)^2

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0\[x^2+2hx+h^2\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3cool so I'm going to put that in place of (x+h)^2 ... \[f(x+h)f(x)=[a(x^2+2xh+h^2)+b(x+h)+c][ax^2+bx+c] \\ \text{ now distributing a bit } \\ f(x+h)f(x)=ax^2+2axh+ah^2+bx+bh+cax^2bxc\] do you have a problem with my distributing step?

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0Nope, easy peasy

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you should see some things that cancel (or zero out) like ax^2ax^2=0 and bxbx=0 and cc=0

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(x+h)f(x)=\cancel{ax^2}+2axh+ah^2+\cancel{bx}+bh+\cancel{c}\cancel{ax^2}\cancel{bx}\cancel{c}\]

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0So we left with \[f(x+h)f(x)=2axh+ah^2+bh\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3now let's go back to our definition this was only the numerator

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)f(x)}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{2axh+ah^2+bh}{h}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so... notice that every term on top has a common factor of h and the denominator also has a common factor h

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f'(x)=\lim_{h \rightarrow 0} \frac{h(2ax+ah+b)}{h(1)}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3we can write as: \[f'(x)=\lim_{h \rightarrow 0} \frac{h}{h} \cdot \frac{2ax+ah+b}{1}\]

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0which the h's cancel, right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3yep and that whole h on the bottom was the thing that was giving us problems but guess what?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3it is no longer there because we canceled that mean h guy on bottom

freckles
 one year ago
Best ResponseYou've already chosen the best response.3so we can plug in 0 now

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0and it leaves us with 2ax+b?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f'(x) =\lim_{h \rightarrow 0} (2ax+ah+b)=2ax+a(0)+b=2ax+0+b=2ax+b\] you are right

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0Is that it?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(x)=ax^2+bx+c \text{ gives us } f'(x)=2ax+b\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3yes now you can find the derivative of any quadratic

freckles
 one year ago
Best ResponseYou've already chosen the best response.3you actually have a formula for it now we just found a formula for it above

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0Awesome! Didn't know it was that simple...the video I watched on it made it seem so confusing lol

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0So there is never a c...

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(x)=ax^2+bx+c \text{ gives } f'(x)=2ax+b \\ \text{ example } \\ \text{ say the question is find } f' \text{ given } f(x)=5x^2+3x4 \\ \text{ we only need to know } a \text{ and } b \\ f'(x)=2(5)x+3=10x+3\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Well the derivative of a constant will always be zero.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Because the slope of y=a constant is zero.

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so very much! :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Np. So this is exactly what you wanted right? Do you have anymore questions about differentiating a quadratic?

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0Yes and no, I don't! :)

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0Wait...perhaps one question...this works for all quadratic equations?

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Alright. I might do one more thing before I leave... A different strategy... Instead of expanding...grouping and factoring... \[f'(x)=\lim_{h \rightarrow 0} \frac{f(x+h)f(x)}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{ a(x+h)^2+b(x+h)+c(ax^2+bx+c)}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{a[(x+h)^2x^2]+b((x+h)x]+[cc]}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{a[(x+h)^2x^2]+b[(x+h)x]+0}{h} \\ f'(x)=\lim_{h \rightarrow 0} \frac{a[(x+hx)(x+h+x)+b[x+hx]}{h} \\ \\ f'(x)=\lim_{h \rightarrow 0} \frac{ah(2x+h)+bh }{h} \\ f'(x)=\lim_{h \rightarrow 0} a(2x+h)+b \\ f'(x)=a(2x+0)+b =2ax+b\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and yes to your question you do know a quadratic function is in this form right: \[f(x)=ax^2+bx+c ?\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3and if you do then yes to your question

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0I love quadratic equations....and finding the solutions using the quadratic formula too :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.3do you know that you can find the xcoordinate of vertex of the parabola using the derivative

freckles
 one year ago
Best ResponseYou've already chosen the best response.3Yep. 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 \[f(x)=ax^2+bx+c \\ f'(x)=2ax+b \\ f'(x)=0 \text{ gives you the equation } 2ax+b=0 \\ \\ \text{ solving for } x \text{ gives you } x=\frac{b}{2a}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3This is exactly what we would get if we did the whole completing the square thing.

freckles
 one year ago
Best ResponseYou've already chosen the best response.3\[f(x)=ax^2+bx+c \\ f(x)=ax^2+\frac{a}{a}bx+c \\ f(x)=a(x^2+\frac{1}{a}bx)+c \\ f(x)=a(x^2+\frac{b}{a}x)+c \\ f(x)=a(x^2+\frac{b}{a}c+(\frac{b}{2a})^2)+ca(\frac{b}{2a})^2 \\ f(x)=(x+\frac{b}{2a})^2+ca(\frac{b}{2a})^2 \\ \text{ where the vertex is } (\frac{b}{2a}, ca(\frac{b}{2a})^2)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.3But anyways all of that is extra.

haleyelizabeth2017
 one year ago
Best ResponseYou've already chosen the best response.0Thank you again :)
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