Question

Consider the functions f(x) = 8x + 16 and g(x) = x^2 + 2x + 5.

Answer 1

Answer:

x =8

Step-by-step explanation:

To answer this question you must find the point at which $g(x)\geq f(x)$

So, we have:

$x^2 + 2x + 5 \geq 8x + 16$

$x^2 + 2x -8x + 5 -16\geq0\\\\x^2 -6x -11\geq 0$

To solve the quadratic function we use the quadratic formula

±

$\frac{-b \± \sqrt{b^2- 4ac}}{2a}$

Where:

$a = 1\\b =-6\\c = -11$

Then:

$\frac{-(-6) \± \sqrt{(-6)^2- 4(1)(-11)}}{2(1)}\\\\x = 7.47\\x = -1.472$

The line cuts the parabola by 2 points, x = -1.472 and x = 7.47.

You can verify that between x = -1.472 and x = 7.47. the line is greater than the parabola, but from x = 7.47, the parabola is always greater than the graph of the line.

Therefore the point sought is:

x = 7.47≈ 8