Question

Find the vertex of the function y = -x2 + x

Answer 1

Just to be clear, is your function:\[y = -x ^{2} +x\]

Answer 2

yes .

Answer 3

what class is this? There are easier ways and harder ways to get the vertex... don't want to give you something you aren't familiar with.

Answer 4

I'd recommend either completing the square of quadratic equation (which are both equivalent)

Answer 5

that's only one option :) equivalent to what?

If it's completing the square, I will happily defer to you...

My other option was finding the min from a derivative and confirming it is vertex (i.e. reflection point), but I am guessing that's too much for this problem

Answer 6

*(typo: ... the square *or* quadratic . . .)

Answer 7

oh, duh, I should have read it that way :)

Answer 8

Finding the derivative is also equivalent to quadratic formula and completing the square, they all yield the same thing. Completing the square is probably the most instructive and more direct for finding the coordinates of the vertex.

Answer 9

Plotting points and graphing can also help build intuition about such things.

Answer 10

@kevin.presley12 do you have a preference?

Answer 11

the anser can either be
1) -3/2; -3
2) 3/2; -3
3) 2/3; -3
4) -2/3; 3

Answer 12

None of those answers work for -x^2+x

Answer 13

Double check to see if you have the right equation and the right answer choices.

Answer 14

i meant 1) (1/2, 1/4)
2) (1/4, 1/2)
3) (1/2, 1/2)
4) (1/4, 1/4)

Answer 15

Ok, I'll show in general how to complete the square:
First, factor out the leading coefficient on the x^2 term.
e.g. given \[ax^2+bx+c \rightarrow a(x^2+\frac{b}{a}x+\frac{c}{a})\]

Answer 16

In your case, you would just factor out a -1.

Now take half of the middle coefficient, square it, and simultaneously add it and subtract it (this is effectively adding zero so it doesn't change the number):
\[a(x^2+\frac{b}{a}x+\frac{c}{a}+(\frac{b}{2a})^2-(\frac{b}{2a})^2)\]
Rearranging:
\[\rightarrow a(x^2+\frac{b}{a}x+(\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2), \]
\[\space x^2+\frac{b}{a}x+(\frac{b}{2a})^2=(x+\frac{b}{2a})^2\]

This is the completed square.

For your example, after factoring out -1, you'll have -(x^2-x). half of the coefficient on the x is a -1/2, which squared = 1/4.

Answer 17

Now we have
\[\rightarrow a((x+\frac{b}{2a})^2+\frac{c}{a}-(\frac{b}{2a})^2)\]
Your equation should now look like
\[-(x^2-x+\frac{1}{4}-\frac{1}{4}) \rightarrow -((x-\frac{1}{2})^2-\frac{1}{4})\]

Answer 18

Now multiply the -1 back in and your final equation in 'vertex form' is
\[\rightarrow y=-(x-\frac{1}{2})^2+\frac{1}{4}\]
x-1/2=0 is the equation for the axis of symmetry; solving for x gives you the x-coordinate of the vertex. The 1/4 is the y-coordinate of the vertex.