How do I find the x-intercepts of a quadratic function in vertex form #(x+4.5)^2-6.25#?

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Answer 1 · 2015-05-24T13:04:10

For a quadratic equation of the form
#y = # and expression in #x#

The x-intercepts occur at the points where #y=0#

Given (#y=#)#((x+4.5)^2-6.25)#
We need to solve
#(x+4.5)^2 -6.25 = 0#

#(x+4.5)^2 = 6.25#

#x+4.5 = +- 2.5#

#x = -4.5 +- 2.5#

The x-intercepts are
#(-2)# and #(-7)#