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

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Answer 1 · 2016-01-01T23:47:51

$color(blue)("Vertex " -> (x,y)-> (2, -2)$ ;
$ycolor(white)(.)_("intercept")=2$

$xcolor(white)(.)_("intercpts") ->$ I have taken you to the point where all you have to do is the final arithmetic

Given: $g(x)=x^2-4x+2$
Tony B

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

Consider the coefficient of $(-4)$ from $-4x$
$color(brown)(x_("vertex") = (-1/2)(-4) = +2)$

so by substitution:

$color(brown)(y_("vertex") =(2)^2-4(2)+2 = -2)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The $y_("intercept")$ is at $x=0$

$color(brown)(=> y_("intercept") = (0)^2-4(0)+2 = 2)$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Not so quick to find the values for $x_("intercept")$ as its factors are non integer values.

Consider standard equation form of: $ax^2+bx+c=0$

and the related $x=(-b+-sqrt(b^2-4ac))/(2a)$

giving: $x=(-(-4) +-sqrt((-4)^2-4(1)(2)))/(2(1))$

$color(green)("I will let you solve that part. Compare your answers to the graph.")$

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