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

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Answer 1 · 2015-11-20T18:55:29

The x-coordinate of the vertex $= - \frac{b}{2 a}$ $=-b/(2a)$

Remember, the general form of a quadratic is:

$a x^{2} + b x + c = 0$ $ax^2+bx+c=0$

x-coordinate of vertex $= \frac{- b}{2 a} = \frac{- (- 1)}{2 \times 1} = \frac{1}{2}$ $=(-b)/(2a)=(-(-1))/(2xx1)=1/2$

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

VERTEX $= (\frac{1}{2}, \frac{7}{4})$ $= (1/2, 7/4)$

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

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

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

The x-intercept is when $x = 0$ $x=0$

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

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