Question

Given F(X)=(2x^(2)+5x-12)/(X+4), What Are The Domain And Range?

Answer 1

Whats this?

Answer 2

If you need help solving equation like this you can always use mathway.com or desmos.com

I used both and got this :

Domain: all real numbers
Range: all real numbers

Answer 3

\[f(x)=\frac{ 2x^2+5x-12 }{x+4 }\]
for domain :
\[x+4\neq 0\]
\[x \neq -4\]
Domain: R-{-4}
for range:
let f(x)=y
\[y=\frac{ 2x^2-5x+12 }{ x+4 }\]
\[yx+4y=2x^2-5x+12\]
\[2x^2-5x-yx+12-4y=0\]
\[2x^2-(5+y)x+(12-4y)=0\]
for x to be real disc.>=0
\[b^2-4ac \ge 0\]
\[[-(5+y)]^2-4(2)(12-4y)\ge 0\]
\[25+10y+y^2-96+32y \ge 0\]
\[y^2+42y-71\ge 0\]
\[y^2+42y+21^2\ge 71+21^2\]
\[(y+21)^2\ge 71+441\]
\[(y+21)^2\ge 512\]
\[\left| y+21 \right|\ge \sqrt{256 \times 2}\]
\[\left| y+21 \right|\ge 16\sqrt{2}\]
\[y+21\le -16\sqrt{2}\]
\[y \le -21 -16\sqrt{2}\]
or
\[y+21\ge 16\sqrt{2}\]
\[y \ge -21+16 \sqrt{2}\]
Range is
\[(-\infty,-21-16\sqrt{2}]U[-21+16\sqrt{2},\infty)\]