How do you use the graph to solve #0=x^2+2x-7#?

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Answer 1 · 2017-09-29T15:13:04

#color(blue)((-1, -8)# & #color(red)((-7, 0)#

graph{x^2+2x-7 [-8.93, 8.85, -8.49, 0.4]}

#x=(-b)/(2a)#

#=(-2)/(2*1)#

#=>x=color(blue)(-1#

And, #y=x^2+2x-7#
#=(-1)^2+2(-1)-7#
#=>y=color(blue)(-8#

Therefore, the minimum point is #color(blue)((-1, -8)#.

To find the #y#-intercept, #x=0#
#(0)^2+2(0)-7#
#=color(red)(-7#

Therefore another solution is #color(red)((-7, 0)#.

Its zeroes will be its #x#-intercepts. So,

#x=((-b)+-sqrt(b^2-4ac))/(2a)#... Quadratic Formula

#x=((-2)+-sqrt(2^2-4*1*(-7)))/(2*1)#

#x=(-2+-sqrt(32))/2#

#x=color(orange)(-1+2sqrt(2)# OR #color(blue)(-1-2sqrt(2)#