Question

suppose f(x)= ax^2 + bx + c where a does not equal 0. Show that the vertex of the graph of f is the point (-b/2a, 4ac-b^2/4a)

Answer 1

wow I have actually done this one before
i wonder if i can find it

Answer 2

please hold...

Answer 3

well i couldn't find it but thats okay! :)

Answer 4

\[y=ax^2+bx+c\]
\[ a \neq 0\]
So we need to put this in the form
\[y=a(x-h)^2+k\]
So that means we need to involve completing the square!
So we have \[y=ax^2+bx+c\]
We need the coefficient of \[x^2\] to be 1.
So we can sorta do something like that, but we want the x term with it so we have
\[y=a(x^2+\frac{b}{a}x)+c\]
How do you feel about this so far?

Answer 5

We can complete the square for the thing inside ( )

Answer 6

\[y=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a \cdot (\frac{b}{2a})^2\]

Answer 7

Whatever you add in you must take out!

Answer 8

We added in and took out \[a \cdot (\frac{b}{2a})^2\]
In other words we did this exact thing:
\[a \cdot (\frac{b}{2a})^2-a \cdot (\frac{b}{2a})^2\]

Answer 9

\[y=a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2)+c-a \cdot (\frac{b}{2a})^2 \]
\[y=a(x+\frac{b}{2a})^2+c-a \cdot \frac{b^2}{4a^2}\]

Answer 10

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

Answer 11

You can write those two last terms as one fraction

Answer 12

You need to find a common denominator!

Answer 13

To do that of course.*

Answer 14

\[y=a(x+\frac{b}{2a})^2+\frac{4a}{4a} \cdot c-\frac{b^2}{4a}\]

Answer 15

I multiply by 1 so now I have the same denominator for the last two terms

Answer 16

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

Answer 17

This is vertex form!

Answer 18

Where \[h=\frac{-b}{2a} \text{ and } k=\frac{4ac-b^2}{4a}\]
and the vertex is (h,k)