Question

Can somebody help me on how to find the maximum and minimums of quadratic functions? Thanks so much.

Answer 1

are you in or have you taken calculus yet?

Answer 2

no. im in algebra 1

Answer 3

ok, I forgot how to do it by algebra, sorry I can't help

Answer 4

its fine, thanks so much

Answer 5

The standard method is completing the square. For example, suppose you have the quadratic \[x^2-2x+4.\] You can rewrite this as \[(x^2-2x + 1) + 3 = (x-1)^2 +3.\] Now we see that \[(x-1)^2\] is always non-negative, so this particular quadratic has a minimum at \[x=1,\] and it has a value of 3 there. If you do this process in general, you will find that the quadratic \[ax^2+bx+c\] has a minimum/maximum at \[x=-\frac{b}{2a}.\] (This is where the vertex of the parabola is if you graph the equation.)

Answer 6

how do you complete the square? Im confused on that part

Answer 7

First suppose you're given a quadratic in the form \[x^2+ax+b.\] We want to rewrite this as \[(x+r)^2\] for some r. Now we know that \[(x+r)^2 = x^2+2rx+r^2.\] So we set \[2r=a \quad \Rightarrow \quad r=\frac{a}{2}.\] But \[(x+a/2)^2 = x^2 + ax + \frac{a^2}{4},\] so we need to add \[b^2-\frac{a^2}{4}\] to both sides to get the original quadratic. So after completing the square our quadratic becomes \[(x+a/2)^2 + b^2-\frac{a^2}{4}.\] Now if you have a quadratic of the form \[ax^2+bx+c\] (where the coefficient of the square term doesn't have to be 1), just factor a from the expression and proceed in the same way for the inside quadratic.

Answer 8

what does the r represent?

Answer 9

It's just a temporary variable I made to represent a constant that is added to x before we square the expression. That was just the form we wanted to rewrite the quadratic in order to complete the square. The first quadratic reaches its minimum when \[x=-r\] because that makes the squared term 0.
Now I should have added that we were trying to rewrite the quadratic \[x^2+ax+b\] in the form \[(x+r)^2 + k\] for some constant k. In the end that's what we did.

Answer 10

the function that i am trying to do this on is f(x)= 3x^2+6x+7. Could you walk me through the problem using these formulas? Im still kind of confused on what the variables represent in the equation

Answer 11

Okay. Let's factor out 3 from the first two terms: \[3(x^2+2x)+7.\] Now I want to make \[(x^2+2x)\] a perfect square by adding a constant to it: \[3(x^2+2x+t)+7.\] However, that just increased the function by 3t (note that there's a 3 outside.) So in order for the expression to remain the same, I have to subtract 3t at the end: \[3(x^2+2x+t)+7 -3t.\] Now we have to choose a t such that \[(x^2+2x+t)\] is a perfect square. What t should we use?

Answer 12

Hint: What is \[(x+1)^2?\]

Answer 13

im not really sure what that means, but (x+1)^2 can be written as (x+1) (x+1), but if t is a perfect square (it has to be to be in a perfect square equation) than it could be something's square root? Im not sure but thats all i got lol

Answer 14

Well t does not always have to be a perfect square, but you're right that it is in this case. Now you said that \[(x+1)^2=(x+1)(x+1),\] and that's right. But can you expand the answer further?

Answer 15

not that i have learned, but whether you can or not is open for discussion

Answer 16

Woh, this whole process wasn't really needed! It's good to know but...

To find the Maximum or Minimum point of a quadratic *also called the vertex*..use the equation that was posted in the very beginning

Your function looks like \(\large y = ax^2 + bx + c\)

The x-coordinate of the vertex can be found by using the formula
\[\large x = -\frac{b}{2a}\]
To then find the y-coordinate, simply plug in this 'x' you just found into the original function

Answer 17

@johnweldon1993 , i got (-1,5), is this correct. If so, is it the minimum?

Answer 18

lol wait i think its (-1,4)

Answer 19

\[\large y= 3x^2 + 6x+ 7\]

So
a = 3
b = 6
c = 7

\[\large x = -\frac{b}{2a} = -\frac{6}{2(3)} = -\frac{6}{6} = -1\]

Good...now plugging that back in you get

\[\large y(-1) = 3(-1)^2 + 6(-1) + 7\]
\[\large y(-1) = 3(1) - 6 + 7\]
\[\large y(-1) = 3 + 1\]
\[\large y(-1) = 4\]

So I have the vertex at (-1,4) indeed

And that is a minimum you say? How do you know? :)

Answer 20

because when you graph it, the lowest point of the parabola is the vertex (-1,4)?

Answer 21

^ @johnweldon1993

Answer 22

Perfect :)

Answer 23

thank you so much for your help! :) @johnweldon1993

Answer 24

Of course :)