How do you find the axis of symmetry, vertex and x intercepts for $y=1/2x^2-x-1$ ?

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Answer 1 · 2016-12-18T22:53:50

Axis of symmetry is $x=1$

$x_("intercepts")=0.73" and "-2.73$ to 2 decimal places

$"Vertex "->(x,y)=(1,-3/2)$

$color(brown)("Note that the question does not state the method to be used.")$

$color(blue)("A quick trick to find the axis of symmetry.")$

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

Using the -2 from $-2x$ we have:

$color(blue)(x_("vertex")=(-1/2)xx(-2) = +1) color(red)(larr"axis of symmetry " x=1)$
The above process is part of completing the square.

$color(blue)(y_("vertex") = 1/2(1)^2-(1)-1 = - 1 1/2->-3/2)$

$color(blue)(=>"Vertex "->(x,y)=(1,-3/2))$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

To find the x-intercepts

Using the formula

$y=ax^2+bx+c" "=>" "x=(-b+-sqrt(b^2-4ac))/(2a)$

$a=1/2; b=-1; c=-1$

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

$x=-1+-sqrt(3)$

$x=0.73205..." and "x=-2.73205...$

Tony B Tony B

How do you find the axis of symmetry, vertex and x intercepts for y=1/2x^2-x-1y=1/2x^2-x-1?