How do you find the vertex and intercepts for g(x) = x^2 - 4x + 2?

1 Answer
Jan 1, 2016

color(blue)("Vertex " -> (x,y)-> (2, -2);
$y {\textcolor{w h i t e}{.}}_{\text{intercept}} = 2$

$x {\textcolor{w h i t e}{.}}_{\text{intercpts}} \to$ I have taken you to the point where all you have to do is the final arithmetic

Explanation:

Given: $g \left(x\right) = {x}^{2} - 4 x + 2$

The quick way for some of it!
There is no coefficient in front of ${x}^{2}$ so we can use:

Consider the coefficient of $\left(- 4\right)$ from $- 4 x$
$\textcolor{b r o w n}{{x}_{\text{vertex}} = \left(- \frac{1}{2}\right) \left(- 4\right) = + 2}$

so by substitution:

$\textcolor{b r o w n}{{y}_{\text{vertex}} = {\left(2\right)}^{2} - 4 \left(2\right) + 2 = - 2}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The ${y}_{\text{intercept}}$ is at $x = 0$

$\textcolor{b r o w n}{\implies {y}_{\text{intercept}} = {\left(0\right)}^{2} - 4 \left(0\right) + 2 = 2}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Not so quick to find the values for ${x}_{\text{intercept}}$ as its factors are non integer values.

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

and the related $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

giving: $x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(2\right)}}{2 \left(1\right)}$

$\textcolor{g r e e n}{\text{I will let you solve that part. Compare your answers to the graph.}}$