How do you solve $- x^{2} + x + 5 > 0$ $-x^2+x+5>0$ 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.

$- x^{2} + x + 5 = 0$ $- x^2 + x + 5 = 0$

using $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{- 2 a}$ $x = ( -b +- sqrt(b^2 - 4ac))/ (-2a)$

$x = \frac{- 1 \pm \sqrt{1^{2} - 4 (- 1) 5}}{- 2 (- 1)}$ $x = ( -1 +- sqrt(1^2 - 4(-1)5))/ (-2(-1))$

$x = \frac{- 1 \pm \sqrt{21}}{2}$ $x = ( -1 +- sqrt(21))/ 2$

$x = - 1.791, 2.791$ $x = -1.791, 2.791$

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

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

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

Hence, -1.791 < x < 2.791

How do you solve −x2+x+5>0-x^2+x+5>0 by graphing?