How do find the vertex and axis of symmetry, and intercepts for a quadratic equation $y=x^2-x-4$ ?

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Answer 1 · 2015-06-06T21:48:06

$y = x^2-x-4 = (x-1/2)^2-1/4-4$

$=(x-1/2)^2-17/4$

The vertex is at $(1/2, -17/4)$

The axis of symmetry is the vertical line $x = 1/2$

The intercept with the $y$ axis is $(0, -4)$

The intercepts with the $x$ axis are the points where $y=0$ , so

$(x-1/2)^2 = 17/4$

$(x-1/2) = +- sqrt(17/4) = +-sqrt(17)/2$

So

$x = 1/2+-sqrt(17)/2 = (1+-sqrt(17))/2$

That is $((1-sqrt(17))/2, 0)$ and $((1+sqrt(17))/2, 0)$

Answer 2 · 2015-06-06T21:52:55

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

y of vertex: $f(1/2) = 1/4 - 1/2 - 4 = -17/4$

Axis of symmetry: x = (-b/2a) = 1/2

x-intercepts: y = x^2 - x - 4 = 0

$D = d^2 = 1 + 16 = 19 -> d = +- sqrt17$

$x = 1/2 +- sqrt17/2$

How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y=x^2-x-4y=x^2-x-4?