Question

For the function F(x) = 2x^2-31x-51, where is the function increasing and positive? ( for all X values greater than what number )

Answer 1

ANSWER
$x \: > \: 17$

EXPLANATION

The given function is

$f(x) = 2 {x}^{2} - 31x - 51$
Let us split the middle term to get,

$f(x) = 2 {x}^{2} -34x + 3x- 51$

We factor to obtain,

$f(x) = 2x(x - 17) -+3(x - 17)$

$f(x) = (2x -+3)(x -17)$

Let us find the zeros of f(x),

$(2x +3)(x - 17) = 0$

Either

$2x +3 = 0 \: or \: x - 17 = 0$
$x = -\frac{3}{2} \: or \: x = 17$

We need to find the x-value of the vertex, which is the midpoint of the x-intercepts.

This gives us,

$\frac{- \frac{3}{2} + 17}{2} = 7.75$
Hence the axis of symmetry of the parabola has equation,

$x = 7.75$

Since the given function has minimum turning point, it will decrease and then increase.

The function will be increasing for
$x \: > \: 7.75$

The function will be increasing and positive for
$x \: > \: 17$