# How do you identify the important parts of y=x^2-4x+2 to graph it?

Jun 17, 2018

See explanation.

#### Explanation:

Given: $y = {x}^{2} - 4 x + 2 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left(1\right)$

$\textcolor{b l u e}{\text{They may not have shown you this trick.}}$

Consider the standardised form $y = a {x}^{2} + b x + c$

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

This is part way to completing the square.

Then ${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{b}{a}$ which in this case is:

${x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{- 4}{1} = + 2$

Then by substitution you can dtermine ${y}_{\text{vertex}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{How they may expect you to do it by completing the square}}$

$\textcolor{b r o w n}{\text{Determine the vertex}}$

Given that $y = a {x}^{2} + b x + c \text{ then } y = a {\left(x + \frac{b}{2 a}\right)}^{2} + c$

But this has introduces a value that is not in the original equation so include a correction. I choose $k$ giving:

$y = a {\left(x + \frac{b}{2 a}\right)}^{2} + k + c$

Set $a {\left(\frac{b}{2 a}\right)}^{2} + k = 0$ neutralising the error.

For this question we have:

$\textcolor{g r e e n}{y = 1 {\left(x + \frac{- 4}{2 \times 1}\right)}^{2} + \textcolor{red}{k} + 2} \text{ } \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(2\right)$

$\textcolor{g r e e n}{\text{Set } {\left(\frac{- 4}{2}\right)}^{2} + \textcolor{red}{k} = 0 \implies \textcolor{red}{k} = - 4}$

$\textcolor{g r e e n}{y = {\left(x - 2\right)}^{2} \textcolor{red}{- 4} + 2}$

$\textcolor{g r e e n}{y = {\left(x \textcolor{b l a c k}{- 2}\right)}^{2} \textcolor{\mathrm{da} r k v i o \le t}{- 2}} \text{ } \ldots \ldots . E q u a t i o n \left(3\right)$

$\textcolor{g r e e n}{{x}_{\text{vertex}} = \left(- 1\right) \times \left(\textcolor{b l a c k}{- 2}\right) = + 2}$
$\textcolor{g r e e n}{{y}_{\text{vertex}} = \textcolor{\mathrm{da} r k v i o \le t}{- 2}}$

$\text{Vertex } \to \left(x , y\right) = \left(2 , - 2\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Determine the y -intercept}}$

Consider the original equation: $y = {x}^{2} - 4 x \textcolor{red}{+ 2}$

${y}_{\text{intercept}} = \textcolor{red}{+ 2}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b r o w n}{\text{Determine the x -intercepts}}$

The ${x}^{2}$ term is positive so the graph is of form $\cup$

${y}_{\text{vertex}} = - 2$ so the graph crosses the x-axis so there are two solution for $y = 0 = {\left(x - 2\right)}^{2} - 2$

$2 = {\left(x - 2\right)}^{2}$
$\pm \sqrt{2} = x - 2$
$x = 2 \pm \sqrt{2}$
$x \approx 3.414213 \ldots . \to x \approx 3.41$ to 2 decimal places
$x \approx 0.585786 \ldots . \to x \approx 0.59$ to 2 decimal places