How do find the vertex and axis of symmetry, and intercepts for a quadratic equation #y= x^2+ 5x - 12#?

1 Answer
George C.
Jun 14, 2015

#y = x^2+5x-12 = (x+5/2)^2-25/4-12#

#= (x+5/2)^2-73/4#

Hence vertex #(-5/2, -73/4)#, axis of symmetry #x=-5/2#, #y# intercept #(0, -12)#, #x# intercepts #(-5/2+-sqrt(73)/2, 0)#

Explanation:

#(x+5/2)^2 = x^2+5x+25/4#

So

#y = x^2+5x-12 = (x+5/2)^2-25/4-12#

#= (x+5/2)^2 - 73/4#

The minimum value of #(x+5/2)^2# occurs when #(x+5/2) = 0#, that is when #x = -5/2#

If #x+5/2 = 0# then #y = -73/4#.

So the vertex is at #(-5/2, -73/4)#

The parabola is vertical, so the axis of symmetry is just the vertical line through the vertex, i.e. #x=-5/2#.

The intercept with the #y# axis occurs where #x=0#. From the original equation we find #y = x^2-5x-12 = 0^2-5*0-12 = -12#. So the #y# intercept is at #(0, -12)#.

The intercepts with the #x# axis are where #y=0#. Then:

#(x+5/2)^2 - 73/4 = 0#

So #(x+5/2)^2 = 73/4#

So #x+5/2 = +-sqrt(73/4) = +-sqrt(73)/2#

and #x = -5/2+-sqrt(73)/2#

That is the #x# intercepts are #(-5/2+-sqrt(73)/2, 0)#

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