Question

help please?
f (x )= -x^2+4x+2
the minimum y-value is y=
@HawkCrimson

Answer 1

okay, so you want to minimize this function. Are you in calculus?

Answer 2

algebra 2

Answer 3

Okay, so you need to find the vertex of the parabola. Do you know how to do that?

Answer 4

wait hold on, -x^2?

Answer 5

Are you sure it's not maximum?

Answer 6

yes -x^2

Answer 7

sorry, yes, maximum, my mistake

Answer 8

-x^2+4x+2=6-(x-2)^2, now since (x-2)^2 is always nonnegative, ....

Answer 9

I come up with -2... but is this right?

Answer 10

Ah, that makes more sense. So, we solve for the x value that gives us the highest value, and then plug it into our equation.

Answer 11

Should be 2

Answer 12

Then plug it into your equation to get 6

Answer 13

can you show me from 2?

Answer 14

so, the +2 at the end is just a vertical shift, so it doesn't matter for finding the maximum x value. So, we just have the equation -x^2 + 4x = 0, and find the zeros. This gives us the two zeros, and the vertex by definition, is halfway between them.

The equation is x(4-x)=0 so our zeros are x=0 and x=4. taking the average value

(4+0)/2 = 2

Plugging this number back into our equation, we get -(2)^2 + 4(2) + 2 = -4+8+2 = 6

Answer 15

Make sense?

Answer 16

ah, I see where I messed up the negative. so 6 is my maximum.

Answer 17

Yep!

Answer 18

thanks for the explanation :)