how does one work out the maximum point by completing the square

osmosis786

2. Originally Posted by osmosis786
how does one work out the maximum point by completing the square

osmosis786
We cant understand your problem, unless you post a problem..

3. Originally Posted by osmosis786
how does one work out the maximum point by completing the square

osmosis786
For the quadratic equation: $ax^2+bx+c = 0$

Divide through by a: $x^2+\frac{b}{a}x + \frac{c}{a} = 0$

Take c/a from both sides: $x^2+\frac{b}{a}x = -\frac{c}{a}$

Then make it into the form (ax+b)^2 and add b^2 to both sides: $(x+\frac{b}{2a})^2 = \frac{b^2}{4a^2} -\frac{c}{a} = \frac{b^2-4ac}{4a^2}$

Can you go from there?

4. ## how do i find the maximum point of this equation using completing the square

how do i find the maximum point of this equation using completing the square

7x + 15 − 2x^2

can u please help as i know how to do minimum points but not maximum points thanks you

5. Originally Posted by osmosis786
how do i find the maximum point of this equation using completing the square

7x + 15 − 2x^2

can u please help as i know how to do minimum points but not maximum points thanks you
???? I thought the method in finding the minimum point or the maximum point of a quadratic is the same -- you put the equation into the form $y = a(x - h)^2 + k$ by completing the square. It's just that when the leading coefficient is positive the vertex is the minimum point, and when the leading coefficient is negative the vertex is the maximum point.

$y = 7x + 15 - 2x^2$
$y = -2x^2 + 7x + 15$
$y = -2\left(x^2 - \frac{7}{2}x\right) + 15$

Now complete the square. When you can get the quadratic into the form
$y = a(x - h)^2 + k$, the vertex is (h, k).

01

6. Originally Posted by yeongil
???? I thought the method in finding the minimum point or the maximum point of a quadratic is the same -- you complete the square. It's just that when the leading coefficient is positive the vertex is the minimum point, and when the leading coefficient is negative the vertex is the maximum point.

$y = 7x + 15 - 2x^2$
$y = -2x^2 + 7x + 15$
$y = -2\left(x^2 - \frac{7}{2}x\right) + 15$

Now complete the square. When you can get the quadratic into the form
$y = a(x - h)^2 + k$, the vertex is (h, k).

01
yes but in the minimum point one must get 'x' to be the smallest possible value so it must be 0

however i do not know how to get 'X' to be as big as possible

maybe i am being stupid and there is a simple thing to do but i cant get my head round it

7. Originally Posted by osmosis786
yes but in the minimum point one must get 'x' to be the smallest possible value so it must be 0

however i do not know how to get 'X' to be as big as possible

maybe i am being stupid and there is a simple thing to do but i cant get my head round it
Sorry, I do not understand what you're saying here. Maybe because of our locations the terminology is different, but I always understood that in a parabola the minimum or maximum point is at the vertex.

01

8. Originally Posted by yeongil
Sorry, I do not understand what you're saying here. Maybe because of our locations the terminology is different, but I always understood that in a parabola the minimum or maximum point is at the vertex.

01
errmmm i think i know how to do it know...

but thanks to everyone who tried to help even though i wasnt being clear ......

9. Originally Posted by osmosis786
yes but in the minimum point one must get 'x' to be the smallest possible value so it must be 0

however i do not know how to get 'X' to be as big as possible

maybe i am being stupid and there is a simple thing to do but i cant get my head round it
You mean y, correct? And you do realize our minimums can be negative (i.e. -5 which is less than 0)?