Vertex of the graph of a quadratic polynomial

This is what a website states:

Before graphing a quadratic function we rearrange the equation, from this:

$f(x) = ax^2 + bx + c$

To this:

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

Where:

$h = -b/2a$
$k = f( h )$

In other words, calculate h$(=-b/2a)$, then find $k$ by calculating the whole equation for $x=h$

The website states the following reason for doing this:

Well, the wonderful thing about this new form is that $h$ and $k$ show us the very lowest (or very highest) point, called the vertex:

And also the curve is symmetrical (mirror image) about the axis that passes through $x=h$, making it easy to graph I want to know that how is $h$ the x - coordinate and $k$ the y - coordinate of the vertex?

The square of any real number is always at least zero, and equals zero only when that number is itself zero. So $(x-h)^2$ is always at least zero, and is equal to zero only when $x-h=0$; that happens only when $x=h$.

Since $(x-h)^2$ is always at least zero, we see that any positive multiple of it is also always at least zero. So, if we are told that $a>0$, we can say with certainty that $a(x-h)^2$ is always at least zero and is equal to zero only when $x=h$.

Finally, we see that $a(x-h)^2 +k$ is always at least $k$ and is equal to $k$ only when $x=h$. That's because we are adding to $k$ a number that is always at least zero (so, the sum can't be any less than $k$).

Clearly $(h,k)$ is a point on the graph of the function $f(x) = a(x-h)^2+k$. Any other point $(x,y)$ on the graph has an $x$-coordinate that is not $h$, so the $y$-coordinate must be greater than $k$. This means that if we move left or right from $(h,k)$, staying on the graph, we can only move up.

Done.

(The case $a<0$ is handled similarly -- try it yourself! What happens if $a=0$?)

Addendum: How to find $h$ and $k$

This is just completing the square to rewrite the original expression. Start with:

$$f(x) = ax^2 + bx + c$$

Now suppose there is and $h$ and a $k$ so that this is the same as

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

Let's write this out in full : $$a(x-h)^2 + k=a(x^2-2hx + h^2) + k = ax^2-2ahx + ah^2+k$$ Now if these are to be the same function for all $x$, then $$ax^2 + bx + c = ax^2-2ahx + ah^2+k$$ Move everything to one side (note the $ax^2$ terms add out): $$(\underbrace{b+2ah}_{\textrm{constant}})x + (\underbrace{c-ah^2-k}_{\textrm{constant}})=0$$

This has to be true for all $x$. But this is just a linear function of $x$, and the only linear function which is zero for all $x$ is the linear function whose coefficients are all zero! This means we can write $$\begin{cases}b+2ah = 0\\c-ah^2-k = 0\end{cases}$$ So solve the first equation for $h$ to get $$\boxed{h=-\tfrac b{2a}}$$ Now you could plug this into the second equation and solve for $k$, but it's probably easier to note that, since we assumed $$f(x) = a(x-h)^2+k$$ then just putting $x=h$ into this we get $$\boxed{f(h)} = a(h-h)^2+k=a\cdot 0^2 + k = \boxed{k}$$ In other words, get $k$ by using the fact that $$\boxed{k=f(h)}$$ (and we know $h$ at this point).

• I got your point that h and k would be coordinates of vertex. But I still didn't understood how we reached the conclusion that h = -b/2a and k = f(h). – Parth Apr 21 '16 at 17:11
• Okay, I added a section in my answer to show how you know this. – MPW Apr 21 '16 at 20:25
• How is b - 2ah = 0? Shouldn't it be b + 2ah? – Parth Apr 22 '16 at 0:49
• @unknownCoder: Yes, typo. Will fix. Thanks for the catch, +1. – MPW Apr 22 '16 at 12:48
• What is the need of including an a in a(x - h)^2 + k? – Parth Apr 22 '16 at 15:13

The quantity $(x-h)^2$ is always non negative, and the smallest value it can attain is zero when $x=h$. Now we have two cases:

1) $a>0$: in this case the quantity $a(x-h)^2$ is always non negative and has zero as its smallest value when $x=h$.

2) $a<0$: in this case the quantity $a(x-h)^2$ is always non positive and has zero as its largest value when $x=h$.

In both cases $x=h$ represents a global minimum or maximum of the function; this property doesn't change if you consider the "shifted" quantity $a(x-h)^2+k$, because adding a constant to a function does not change the position of its maxima or minima.

So the x-coordinate of the vertex is at $x=h$.

(1) All graphs of quadratic functions are symmetric about some vertical axis. $y = f(x) = ax^2 + bx + c$ (where we assume that $a > 0$ for simplicity) is of no exception.

(2) The vertex $V(h, k)$ will lie on that axis.

(3) Let us assume that $f(x) = 0$ has roots $\alpha$ and $\beta$. Then, the graph of y = f(x) will cut the x-axis at $A( \alpha, 0)$ and $B( \beta, 0)$. The point $( \dfrac {\alpha + \beta}{2}, 0)$ is midway between A and B. Then, the vertical axis (of symmetry) has the equation $L : x = \dfrac {\alpha + \beta}{2}$.

(4) Recall that $\alpha + \beta =$ the sum of roots $= \dfrac {–b}{a}$. Therefore, equation of L is then $L : x = \dfrac {–b}{2a}$.

(5) According to (2), $V = (h, k) = (\dfrac {–b}{2a}, k)$; where k is given by $f(\dfrac {–b}{2a})$ or $f(h)$ because $V$ is a point on the curve of $y = f(x)$.

In the case that $f(x) = 0$ has no real roots, (ie. the graph of $y = f(x)$ does not cut the x-axis), we can translate the x-axis upward certain suitable units such that the new X-axis will cut $y = f(x)$. The above argument still applies.