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

Not quite sure how to factorise this one...

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

Any input greatly appreciated.

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

2. Originally Posted by davidman
Not quite sure how to factorise this one...

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

Any input greatly appreciated.

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

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

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

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

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

3. Originally Posted by davidman
Not quite sure how to factorise this one...

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

Any input greatly appreciated.

So far, I tried $\displaystyle y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9\times2-5=-2(x-3)^2+13$
hi
$\displaystyle y=-(4x^2-12x+5)=-[(2x-3)^2-9+5]=-(2x-3)^2+4.$

4. Originally Posted by davidman
Not quite sure how to factorise this one...

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

Any input greatly appreciated.

So far, I tried $\displaystyle y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9-\frac{5}{2}$
Factor out the leading coefficient from the first two terms

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

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

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

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

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

Now factor the trinomial square, and combine like terms

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

5. Originally Posted by Raoh
hi
$\displaystyle y=-(4x^2-12x+5)=-[(2x-3)^2-9+5]=-(2x-3)^2+4.$
$\displaystyle -(2x-3)^2+4=-(2(x-\frac{3}{2}))^2+4=-4(x-\frac{3}{2})^2+4$.

6. [quote=VonNemo19;442494]

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

[\quote]

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

7. [QUOTE=Masterthief1324;442892]
Originally Posted by VonNemo19

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

[\quote]

Complete the square and you get $\displaystyle \frac {9}{4}$ so where did the $\displaystyle \frac {-9}{4}$ 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 $\displaystyle -4$ to the last term in the parentheses and "kicked it out" of the parentheses.

Example:

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

8. [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 $\displaystyle -4$ to the last term in the parentheses and "kicked it out" of the parentheses.

Example:

$\displaystyle 3(x^2+2x)=3(x^2+2x+1-1)=3(x^2+2x+1)+3(-1)=3(x^2+2x+1)-3=3(x+1)^2-3$
I don't quite understand. If it isn't a burden, can you refer me to an article that covers this?

9. [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.

10. Originally Posted by davidman
I'm looking for the turning point of $\displaystyle y=-4x^2+12x-5$ so I want $\displaystyle y$ in the form $\displaystyle y=a(x-p)^2+q$.
You can also find the turning-point (of a quadratic equation) without completing the square. It's at $\displaystyle \left(\frac{-b}{2a}, -\frac{\left(b^2-4ac\right)}{4a}\right)$. So for your equation, for example, with $\displaystyle a = -4, b = 12,$ and $\displaystyle c = -5$, we have:

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