Question

i dont understand deriving the vertex formula its very confusing...help?

Answer 1

\[y=ax^2+bx+c\] vertex is
\[x=-\frac{b}{2a}\]

Answer 2

vertex for which equation?

Answer 3

and y is what you get when you plug it in for x

Answer 4

vertex is the stationary point so when dy/dx=0

Answer 5

lol

Answer 6

or when y=ax^2+bx+c
dy/dx=2ax+b=0
x=-b/2a

Answer 7

so for example if you have
\[y=x^2-6x+2\] then
\[-\frac{b}{2a}=\frac{6}{2}=3\]

Answer 8

and the second coordinate would be
\[3^2-6\times 3+2=9-18+2=-9+2=-7\] and so vertex would be
\[(3,-7)\]

Answer 9

it is where the parabola "sits" so if the leading coefficient is positive then the parabola faces up and the vertex gives the lowest point on the curve. if the leading coefficient is negative then the vertex gives the highest point

Answer 10

you need it "derived" as in how do you know that is the vertex?

Answer 11

\[y=ax^2+bx+c=ax^2+\frac{a}{a}bx+c=a(x^2+\frac{1}{a}bx)+c\]
\[=a(x^2+\frac{b}{a}x)+c=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2\]
\[=a(x+\frac{b}{2a})^2+c-a \frac{b^2}{4a^2}=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}\]
\[=a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}\]

Answer 12

reason as follows
\[y=ax^2+bx+c\]
\[y=a(x+\frac{b}{2a})^2+\text{stuff}\]
first term is always positive (resp negative) if a is positive (resp negative) so smallest (biggest) it can be is zero, and it is zero when
\[x=-\frac{b}{2a}\]
done

Answer 13

oh my myininaya there you go again with that formula for finding the y value when
\[x=-\frac{b}{2a}\] i know you like that one

Answer 14

vertex is \[(-\frac{b}{2a},\frac{4ac-b^2}{4a})\]

now we can verify the y-coordinate another way by plugging in -b/2a into original function
\[f(x)=ax^2+bx+c\]

\[f(\frac{-b}{2a})=a(\frac{-b}{2a})^2+b(\frac{-b}{2a})+c\]
\[=a \frac{b^2}{4a^2}+\frac{-b^2}{2a}+c=\frac{b^2}{4a}+\frac{-b^2}{2a}+c\]
\[=\frac{b^2}{4a}+\frac{2}{2}\frac{-b^2}{2a}+\frac{4a}{4a}c\]
\[=\frac{b^2-2b^2+4ac}{4a}=\frac{-b^2+4ac}{4a}=\frac{4ac-b^2}{4a}\]

Answer 15

any questions? :)

Answer 16

satellite you will learn to like my way(s)

Answer 17

yes i have a question. why would i bother to use
\[\frac{4ac-b^2}{4a}\] when i can just plug in
\[-\frac{b}{2a}\]?

Answer 18

$$ax^2+bx+c=0$$

roots are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

vertex is located in the middle between the two roots

so
$$\frac{1}{2}\left(\frac{-b-\sqrt{b^2-4ac}}{2a}+\frac{-b+\sqrt{b^2-4ac}}{2a}\right)$$

$$=\frac{1}{2}\left(\frac{-b-\sqrt{b^2-4ac}-b+\sqrt{b^2-4ac}}{2a}\right)$$

$$=\frac{1}{2}\left(\frac{-2b}{2a}\right)$$
$$=\frac{-b}{2a}$$

Answer 19

satellite i'm just showing what you get when you plug it in lol
nice zarkon