# How do you find the important points to graph F(x)= 2x^2 - 4x +1?

Nov 1, 2015

See explanatioin

#### Explanation:

$\textcolor{g r e e n}{\text{Standard form for a quadratic is: } a {x}^{2} + b x + c}$

$\textcolor{g r e e n}{\text{Convert this into } a \left({x}^{2} + \frac{b}{a} + \frac{c}{a}\right)}$

$\textcolor{g r e e n}{\text{In your case we factorise it into}}$

$\textcolor{g r e e n}{y = 2 \left({x}^{2} - 2 x + \frac{1}{2}\right)}$

$\textcolor{g r e e n}{\text{It is important that there is no coefficient directly in front of the } x}$

In this case we look at what is inside the brackets to get what we want.

${x}^{2}$ is positive: that gives you an upwards horse shoe shape.
$\text{ }$ If it had been negative then the horse shoe would
$\text{ }$ have been the other way up.
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$\textcolor{b l u e}{\text{The maxima/minima x value is } \left(- \frac{1}{2}\right) \times \frac{b}{c}}$

So for you it will be at $\textcolor{red}{{x}_{\text{minima}}} = \left(- \frac{1}{2}\right) \times \left(- 2\right) = \textcolor{red}{1}$
To find the y value at the minima substitute $\left(- \frac{1}{2}\right) \times \left(- 2\right) = \left(1\right)$ for x

So
$y = 2 {x}^{2} - 4 x + 1$
becomes:
$\textcolor{red}{{y}_{\text{minima}}} = 2 {\left(1\right)}^{2} - 4 \left(1\right) + 1 = \textcolor{red}{- 1}$

Thus:

$\textcolor{red}{{\left(x , y\right)}_{\text{minima}}}$ $\textcolor{b l u e}{\to \left(1 , - 1\right)}$
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$\textcolor{b l u e}{\text{To find y intercept substitute " x = 0" and solve}}$

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$\textcolor{b l u e}{\text{To find x intercept substitute "y=0" and solve}}$

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