Given $f(x)=x^{2}+4 x+15$ find all values for $x$ for which $f(x) \leq 3$.
A) $(-\infty,-6] \cup[2, \infty)$
B) $(-6,2)$
C) $[-6,2]$
D) $\varnothing$

Question

Quizplus · Accepted Answer

To find the values of $x$ for which $f(x) \leq 3$, we first set $f(x) = 3$ and solve for $x$. The equation becomes $x^2 + 4x + 15 = 3$, simplifying to $x^2 + 4x + 12 = 0$. The discriminant of this quadratic equation is $b^2 - 4ac = 4^2 - 4(1)(12) = 16 - 48 = -32$, which is less than 0. This means the equation has no real solutions, and $f(x) > 3$ for all real $x$. Therefore, there are no values of $x$ for which $f(x) \leq 3$, making the correct answer $\varnothing$.

Given f(x)=x2+4x+15f(x)=x^{2}+4 x+15f(x)=x2+4x+15 Find All Values For xxx For Which f(x)≤3f(x) \leq 3f(x)≤3

Given $f(x)=x^{2}+4 x+15$ Find All Values For $x$ For Which $f(x) \leq 3$