Question

Find the maximum or minimum of the following quadratic function: y =x2 - 2x - 48.

I solved and the result was -49, may you explain whether or not the answer I found was correct?

Answer 1

Check your section.

Answer 2

Dumb, y = x2 - 2x - 42

try differentiating it both sides..

Answer 3

Why is this in OS Feedback? ;-;

Answer 4

oh sorr

Answer 5

dy/dx = 2x - 2

2x - 2 = 0

x = 1

x = 1 is a critical point.

d2y/dx2 = 2

therefore, x = 1; y = 1 - 2 - 48 = -49 ; minima.

no global maxima exists.

Answer 6

\[\begin{align}y & = x^2 - 2x - 48 \\ &= x^2 - 2x+1-49 \\ & = (x - 1)^2 - 49\end{align}\]A term that goes like \(\rm (something)^2\) must always be greater than or equal to zero. The minimum of this function would be when this term is exactly zero. If this term is exactly zero, then your expression would turn out to be \(0 - 49 = -49\) which is your minimum.

Answer 7

that can be further treated and we can calculate the vertex : (1,-49)

Answer 8

I'm sorry, I hadn't noticed which section I had posted this in!!!
Sorry @Compassionate.

Answer 9

This is the first time I've made this mistake!!

Answer 10

this is showing in mathematics group.. lol! OpenStudy rocks .. haha

Answer 11

Haha! I was pretty sure that I posted it in Math... whatever, maybe a glitch!

Answer 12

Could also find where \(y`=0\), but this works:
https://www.desmos.com/calculator/rotqwuepoy

Answer 13

you mean to use calculator for it eric?

Answer 14

Thank you folks. I'm not sure whether or not to give mathslover or parthkohli the medal, hah!

Answer 15

That graphs is and shows the vertex, which is the point he found. \(y`=0\) is a calculus thing.

Answer 16

Yeah, Eric, that is what the vertex which was found by Parth from (x-1)^2 - 49,
(1,-49)

graphing is nice though. :)

@dumbsearch2 i gave the medal to parth, parth give medal to Eric. :)

Answer 17

I do not really need medals. Hehe

Answer 18

Eric, I was doing a formality .. lol! :)

Answer 19

so the answer is -49

Answer 20

This method is called completing the square. You can derive the maximum/minimum of ANY quadratic function using this.\[\begin{align}y& = ax^2 + bx + c \\ & = a\left(x^2 + \dfrac{b}{a}x + \dfrac{c}{a}\right) \\& = a \left(x^2 + \dfrac{b^2}{(2a)^2} - \dfrac{b^2}{(2a)^2} +\dfrac{b}{a} + \dfrac{c}{a}\right) \\ & = a\left(\left(x + \dfrac{b}{2a}\right)^2- \dfrac{b^2}{4a^2} + \dfrac{c}{a}\right)\\ &= a\left(x + \dfrac{b}{2a}\right)^2 - \dfrac{b^2-4ac}{4a}\end{align}\]When the square-term is zero, the whole function would evaluate to\[\dfrac{-b^2 - 4ac}{4a}~ ~ \text{or more eloquently} -\dfrac{D}{4a}\]Here, \(D\) is the discriminant which you can notice in the quadratic formula.

Answer 21

Yeah, as there is no maximum, so it is -49 (global minimum.)

Answer 22

Note that, I'm referring to \(\bf{\text{No global maximum}}\) . there may be many local maxima. :)

Answer 23

Yeah, I know about completing the square.

Thank you so much @ParthKohli for your in-depth explanation.

Answer 24

Notice one thing more, calculus method generally works while completing the square can not work always.

Answer 25

Damn I'm getting further and further down the hole of confusion :p

Answer 26

Well, completing the square works generally, but it may lead you to a bit confusing in between.

Answer 27

\(y =x^2 - 2x - 48\implies \)
\(y' =2x - 2\implies \)

\(2x - 2=0\implies \)
\(x =1 \) is a critical point where there is a min or max.

Plugging in \(x=1\) into the original gets the -49, then there are some other tests you can use to find if it is a min or a max.

That would be the calculus way. It is overkill for this, where finding the vertex will do it.

Answer 28

A "vertex" is a point which is on the top or the bottom of a graph of a quadratic equation (which is shaped like a parabola).

Consider a U-shaped parabola. If you notice, the vertex is the point where the y-value is minimum, or in other words, the function is minimum. We just found out that the function is minimum at \(x = -\dfrac{b}{2a}\) and \(y = -\dfrac{D}{4a}\). You can memorize these results.