How do you find the vertex and intercepts for $y = x^2 - x + 2$ ?

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Answer 1 · 2015-11-20T18:55:29

The x-coordinate of the vertex $=-b/(2a)$

Remember, the general form of a quadratic is:

$ax^2+bx+c=0$

x-coordinate of vertex $=(-b)/(2a)=(-(-1))/(2xx1)=1/2$

So, when $x=1/2$ , then $y=(1/2)^2-(1/2)+2=7/4$

VERTEX $= (1/2, 7/4)$

Solve for $y=0$ to find the y-intercepts :

$(x-2)(x+1)=0$

So, $x=2$ and $x=-1$

The x-intercept is when $x=0$

$y=(0)^2-(0)+2=2$

How do you find the vertex and intercepts for y = x^2 - x + 2y = x^2 - x + 2?