# For which real x-values lays the graph with equation y = -x^4 + 18x^2 - 17 under the x-axis? Thank you!

Nov 7, 2017

]-oo, -sqrt(17)[uu]-1,1[uu]sqrt(17), +oo[

#### Explanation:

1)You must find the zeros of the equation

This is a second-degree equation where the "x" is ${x}^{2}$

${x}^{2} = \frac{- 18 \pm \sqrt{{18}^{2} - 4 \cdot \left(- 1\right) \left(- 17\right)}}{2 \cdot \left(- 1\right)}$

${x}^{2} = \frac{- 18 \pm \sqrt{324 - 68}}{- 2}$

${x}^{2} = \frac{- 18 \pm \sqrt{256}}{- 2}$

${x}^{2} = \frac{- 18 \pm 16}{-} 2$

${x}^{2} = 9 \pm 8$

${x}^{2} = 1 \mathmr{and} {x}^{2} = 17$

Therefore the roots are:

$x = \pm 1 \mathmr{and} x = \pm \sqrt{17}$

Now we need to know in which direction the polynomium goes in the zeros For this we need the first derivative:

$- 4 {x}^{3} + 36 x$

$- \sqrt{17} \to 131 < 0 \to \uparrow$

$- 1 \to - 32 < 0 \to \downarrow$

$1 \to 32 < 0 \to \uparrow$

$\sqrt{17} \to < - 131 \to 0 \to \downarrow$

Therefore the polynomial grows in -sqrt(17); decresases in -1; groes in 1 and decreases in sqrt(17) (it gives a form of a M). Taking the obtained zeros we can say that its negative in:

]-oo, -sqrt(17)[uu]-1,1[uu]sqrt(17), +oo[