# Math Help - Peak of graph

1. ## [Solved] Peak of graph

I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like

$y=(x+p)^2+q$

is this correct? and how would you do for

$y=2x^2-4x+a^2-a$

2. Originally Posted by davidman
I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like

$y=(x+p)^2+q$

is this correct? and how would you do for

$y=2x^2-4x+a^2-a$

If you write the quadratic equation as

$y = a(x - h)^2 + k$

then the turning point is $(x, y) = (h, k)$.

So for $y = 2x^2 - 4x + a^2 - a$, you complete the square

$y = 2\left(x^2 - 2x + \frac{1}{2}a^2 - \frac{1}{2}a\right)$

$y = 2\left[x^2 - 2x + (-1)^2 - (-1)^2 + \frac{1}{2}a^2 - \frac{1}{2}a\right]$

$y = 2\left[(x - 1)^2 + \frac{1}{2}a^2 - \frac{1}{2}a - 1\right]$

$y = 2(x - 1)^2 + a^2 - a - 2$.

So the turning point is:

$(x, y) = (1, a^2 - a - 2)$.

3. wooah, you made that so easy to understand.

thanks!

4. Originally Posted by davidman
I read somewhere that you can get the coordinates for the peak of a quadratic equation by just rearranging the equation. I forgot how it was done though. Something like

$y=(x+p)^2+q$

is this correct? and how would you do for

$y=2x^2-4x+a^2-a$

If the vertex of the parabola is V(p,q) then the equation of the parabola whose axis is parallel to the y-axis is written as:

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

$y = 2x^2-4x+a^2-a = 2(x^2-2x+\tfrac12(a^2-a))$

$y=2(x^2-2x\bold{\color{red}+1-1}+\tfrac12(a^2-a)) = 2(x-1)^2+a^2-a-2$

So the coordinates of the vertex are $V\left(1, a^2-a-2\right)$

5. Thanks so much for the help!

I've been trying my hand at the second part of this question, but I can't seem to get to the answer shown in my book...

the second part is to get the minima of $f(a)=a^2-a-2$

and I thought the answer would reveal itself if I factorised and solved for $a$, but the answer is apparently $-\frac{9}{4}$... I was not even close... what exactly are you supposed to do to get to that answer?

6. Originally Posted by davidman
Thanks so much for the help!

I've been trying my hand at the second part of this question, but I can't seem to get to the answer shown in my book...

the second part is to get the minima of $f(a)=a^2-a-2$

and I thought the answer would reveal itself if I factorised and solved for $a$, but the answer is apparently $-\frac{9}{4}$... I was not even close... what exactly are you supposed to do to get to that answer?
Probably you are trapped by the text of the question(?) It reads " ... get the minima of f(a) ..."

I assume that you have calculated:

$f(a)=a^2-a-2$

$f(a)=a^2-a\bold{\color{blue}+\frac14-\frac14}-2$

$f(a)=\left(a-\frac12 \right)^2-\frac14-2$

$f(a)=\left(a-\frac12 \right)^2-\frac94$

Therefore the vertex has the coordinates $V\left(\tfrac12 ,~ -\tfrac94 \right)$

With a parabola opening up the f(a)-value of the vertex determines the minimum. The vertex is the lowest point of all points of the graph of f. Therefore

$\min(f(a)) = -\frac94$

7. Thanks a lot guys. earboth's post brings that question to perfection!