What is the vertex of y= x^2 - 4x + 1?

Feb 19, 2016

$\left(2 , - 3\right)$

Explanation:

We start off with $y = {x}^{2} - 4 x + 1$. Now, if we are trying to find a vertex, the easiest way to do that (save for graphing), is to convert your equation into vertex form. And, because I like easier, more time-saving routes, I'll do that.

How we do that is by completing the square.

The first step is to get the coefficient of ${x}^{2}$ equal to $1$. And would you look at that, it already is. Easy for us, right?

Now we have the second step. Something you should know is that the goal of completing the square is to make the ${x}^{2} - 4$ part of our equation into a perfect square. To do that, we take the middle term of the equation, $- 4 x$, and we only look at the coefficient, or the the $- 4$. Now we take the middle term's coefficient, and we divide it in half: $\frac{- 4}{2}$, which equals $- 2$. After that, we square our quotient, like so: ${\left(- 2\right)}^{2}$. This means $\left(- 2\right) \cdot \left(- 2\right)$, or $4$.

So now we have the value that will make ${x}^{2} - 4 x$ a perfect square: $4$! Let's put it together.
$y = {x}^{2} - 4 x + 4 \textcolor{red}{- 4} + 1$
PLEASE NOTE I added and then subtracted the $4$. We have to do this because adding a random number into an equation is, math terms, illegal. We have to add it and then subtract it, because that means we haven't actually changed the value of the equation, we've just sort of "rearranged it".

Now, we can condense $y = {x}^{2} - 4 x + 4$ into $y = {\left(x - 2\right)}^{2}$, and $- 4 + 1$ becomes $- 3$. All together, we now have $y = {\left(x - 2\right)}^{2} - 3$.

From here, finding the vertex is very simple: take the value in the parentheses $\left(- 2\right)$, and reverse the sign, so that now you have $2$. Congratulations, you just found your $x$-value of the coordinate pair.

To find the $y$-value, just take the value of the constant, and don't change any signs; just leave it alone. Now you have all you need to create a coordinate pair.

$\left(2 , - 3\right)$ is the vertex, and lest's just confirm that by graphing the original equation.
graph{y=x^2-4x+1}

And the vertex is $\left(2 , - 3\right)$! Great job!

Feb 19, 2016

Bit of allowed shortcut that relies on the principles behind completing the square. I will let you work out ${y}_{\text{vertex}}$

" " color(blue)(x_("vertex") = 2)

Explanation:

Consider the standard form of:$\text{ } a {x}^{2} + b x + c = 0$

Write this as:$\text{ } a \left({x}^{2} + \frac{b}{a} x\right) + c = 0$

In your question $a = + 1$

now apply:" "(-1/2)xxb/a = x_("vertex")

Then substitute ${x}_{\text{vertex}}$ back into the original equation to find ${y}_{\text{vertex}}$
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$\textcolor{b l u e}{\text{Applying this to your question}}$

Your $\frac{b}{a} \to \left(- \frac{4}{1}\right)$

so $\left(- \frac{1}{2}\right) \times \frac{b}{a} \to \left(- \frac{1}{2}\right) \times \left(- 4\right) = + 2$

So " " color(blue)(x_("vertex") = 2)

$\textcolor{b r o w n}{\text{This matches the graph!}}$

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With practise you could work this bit out in your head. As long as the equation is in the format I showed you.

I will let you work out the value for ${y}_{\text{vertex}}$