How do you find the x and y intercept for $y = x^{2} - 4 x + 4$ $y = x^2 - 4x + 4$ ?

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Answer 1 · 2016-05-08T21:31:43

$x -$ $x-$ intercept is at 2. and $c = 4$ $c = 4$ , the $(y -$ $(y-$ intercept)

To find the $y -$ $y-$ intercept is really easy ... it is the constant.
It is found by making $x = 0$ $x = 0" "$ So, $c = 4$ $c = 4$

To find the $x -$ $x-$ intercept, make $y = 0$ $y = 0$ and solve the equation.

This is the equation of a parabola,
$\Rightarrow$ $rArr$ there could be two, one or no $x -$ $x-$ intercepts.

$x^{2} - 4 x + 4 = 0$ $x^2 - 4x + 4 = 0$

This factorises: $(x - 2) (x - 2) = 0$ $(x-2)(x - 2) = 0$

Although this gives two answers, they are the same: $x = 2$ $x= 2$ ,

so there is actually one intercept on the $x -$ $x-$ axis, which means that the intercept is also the vertex (turning point) of the graph.

How do you find the x and y intercept for y=x2−4x+4y = x^2 - 4x + 4?