How do you solve #-x^2+x+5>0# by graphing?

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Answer 1 · 2016-11-21T03:39:42

-1.791 < x < 2.791
graph{-x^2+x +5 [-10, 10, -5, 5]}

Find the points of intersection of the graph with the x-axis.

# - x^2 + x + 5 = 0 #

using # x = ( -b +- sqrt(b^2 - 4ac))/ (-2a)#

# x = ( -1 +- sqrt(1^2 - 4(-1)5))/ (-2(-1))#

# x = ( -1 +- sqrt(21))/ 2#

# x = -1.791, 2.791#

The point of intersection of the graph with y-axis
is # y = 5# by substituting # x = 0 # into #- x^2 + x + 5 #

Since the coefficient of #x^2# is negative, the curve upwards, as shown in the graph.

For # - x^2 + x + 5 > 0 #, we look at the region of the graph where the curve is above the y-axis.

Hence, -1.791 < x < 2.791