Mathematics: 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 is 43 degrees for #6, and selection C for #7...

For the function f(x) = 2x^2-31x-51, where is - id223151205377, angelasnipes51orfryq

Ответ:

braydenmcd02

10.03.2022

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)

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
$x \: \: 17$

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

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