Answer:
x =8
Step-by-step explanation:
To answer this question you must find the point at which [tex]g(x)\geq f(x)[/tex]
So, we have:
[tex]x^2 + 2x + 5 \geq 8x + 16[/tex]
[tex]x^2 + 2x -8x + 5 -16\geq0\\\\x^2 -6x -11\geq 0[/tex]
To solve the quadratic function we use the quadratic formula
±
[tex]\frac{-b \± \sqrt{b^2- 4ac}}{2a}[/tex]
Where:
[tex]a = 1\\b =-6\\c = -11[/tex]
Then:
[tex]\frac{-(-6) \± \sqrt{(-6)^2- 4(1)(-11)}}{2(1)}\\\\x = 7.47\\x = -1.472[/tex]
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
