y=a(x-p)^2+q

**davidman** · January 21st 2010, 06:23 AM

Not quite sure how to factorise this one...

I'm looking for the turning point of $y=-4x^2+12x-5$ so I want $y$ in the form $y=a(x-p)^2+q$ .

Any input greatly appreciated.

So far, I tried $y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9\times2-5=-2(x-3)^2+13$

**mathaddict** · January 21st 2010, 06:42 AM

Originally Posted by davidman

Not quite sure how to factorise this one...

I'm looking for the turning point of $y=-4x^2+12x-5$ so I want $y$ in the form $y=a(x-p)^2+q$ .

Any input greatly appreciated.

So far, I tried $y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9\times2-5=-2(x-3)^2+13$

hi

$y=-4[x^2-3x]-5$

$=-4[(x-\frac{3}{2})^2-(\frac{3}{2})^2]-5$

$=-4(x-\frac{3}{2})^2+4$

since the leading coefficient is negative , this graph would be a parabola opening downwards , hence the maximum value is 4 , and it occurs when x=3/2

**Raoh** · January 21st 2010, 06:44 AM

Originally Posted by davidman

Not quite sure how to factorise this one...

I'm looking for the turning point of $y=-4x^2+12x-5$ so I want $y$ in the form $y=a(x-p)^2+q$ .

Any input greatly appreciated.

So far, I tried $y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9\times2-5=-2(x-3)^2+13$

hi
$y=-(4x^2-12x+5)=-[(2x-3)^2-9+5]=-(2x-3)^2+4.$

**VonNemo19** · January 21st 2010, 06:45 AM

Originally Posted by davidman

Not quite sure how to factorise this one...

I'm looking for the turning point of $y=-4x^2+12x-5$ so I want $y$ in the form $y=a(x-p)^2+q$ .

Any input greatly appreciated.

So far, I tried $y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9-\frac{5}{2}$

Factor out the leading coefficient from the first two terms

$y=-4(x^2-3x)-5$

Now complete the square. This involve adding and subtracting by the same quantity.

$y=-4(x^2-3x+\frac{9}{4}-\frac{9}{4})-5$

Now distribute the leading coefficient to the last term in the parentheses as to "kick it out"

$y=-4(x^2-3x+\frac{9}{4})+9-5$

Now factor the trinomial square, and combine like terms

$y=-4(x-\frac{3}{2})^2+4$

**Raoh** · January 21st 2010, 06:55 AM

Originally Posted by Raoh

hi
$y=-(4x^2-12x+5)=-[(2x-3)^2-9+5]=-(2x-3)^2+4.$

$-(2x-3)^2+4=-(2(x-\frac{3}{2}))^2+4=-4(x-\frac{3}{2})^2+4$ .
i didn't read your question well

.

**Masterthief1324** · January 21st 2010, 06:00 PM

[quote=VonNemo19;442494]

Now distribute the leading coefficient to the last term in the parentheses as to "kick it out"
$y=-4(x^2-3x+\frac{9}{4})+9-5$

[\quote]

Complete the square and you get $\frac {9}{4}$ so where did the $\frac {-9}{4}$ come from? Any the 9?

**VonNemo19** · January 30th 2010, 10:46 PM

[QUOTE=Masterthief1324;442892]

Originally Posted by VonNemo19

Now distribute the leading coefficient to the last term in the parentheses as to "kick it out"
$y=-4(x^2-3x+\frac{9}{4})+9-5$

[\quote]

Complete the square and you get $\frac {9}{4}$ so where did the $\frac {-9}{4}$ come from? Any the 9?

Note that when add a quantity to an expression, you must also subtract the same quantity as to maintain equality. I distributed the $-4$ to the last term in the parentheses and "kicked it out" of the parentheses.

Example:

$3(x^2+2x)=3(x^2+2x+1-1)=3(x^2+2x+1)+3(-1)=3(x^2+2x+1)-3=3(x+1)^2-3$

**Masterthief1324** · January 31st 2010, 04:01 PM

[quote=VonNemo19;448426]

Originally Posted by Masterthief1324

Note that when add a quantity to an expression, you must also subtract the same quantity as to maintain equality. I distributed the $-4$ to the last term in the parentheses and "kicked it out" of the parentheses.

Example:

$3(x^2+2x)=3(x^2+2x+1-1)=3(x^2+2x+1)+3(-1)=3(x^2+2x+1)-3=3(x+1)^2-3$

I don't quite understand. If it isn't a burden, can you refer me to an article that covers this?

**VonNemo19** · January 31st 2010, 04:05 PM

[QUOTE=Masterthief1324;448941]

Originally Posted by VonNemo19

I don't quite understand. If it isn't a burden, can you refer me to an article that covers this?

Completing the Square: Solving Quadratic Equations

Standard and vertex form of the equation of parabola and how it relates to a parabola's graph.

**Elvis** · January 31st 2010, 04:25 PM

Originally Posted by davidman

I'm looking for the turning point of $y=-4x^2+12x-5$ so I want $y$ in the form $y=a(x-p)^2+q$ .

You can also find the turning-point (of a quadratic equation) without completing the square. It's at $\left(\frac{-b}{2a}, -\frac{\left(b^2-4ac\right)}{4a}\right)$ . So for your equation, for example, with $a = -4, b = 12,$ and $c = -5$ , we have:

$\left(\frac{-b}{2a}, -\frac{\left(b^2-4ac\right)}{4a}\right)$ = $\left(\frac{-12}{2(-4)}, -\frac{\left[(12)^2-4(-4)(-5)\right]}{4(-4)}\right)$ = $\left(\frac{6}{4}, -\frac{(144-80)}{-16}\right)$ = $\left(\frac{3}{2}, \frac{64}{16}\right) = \left(\frac{3}{2}, 4\right).$

Thread: y=a(x-p)^2+q

LinkBack

Thread Tools

Display

y=a(x-p)^2+q

Search Tags