Solve the problem.
-Given $f ( x ) = x ^ { 2 } - 18 x + 68$, find all values for $x$ for which $f ( x ) \geq 0$.
A) $( - \infty , 9 - \sqrt { 13 } ] \cup [ 9 + \sqrt { 13 } , \infty )$
B) $( - \infty , 9 - \sqrt { 13 } ) \cup ( 9 + \sqrt { 13 } , \infty )$
C) $[ 9 - \sqrt { 13 } , 9 + \sqrt { 13 } ]$
D) $( - \infty , - \sqrt { 13 } ] \cup [ \sqrt { 13 } , \infty )$

Question

Quizplus · Accepted Answer

To find the values of $x$ for which $f(x) \geq 0$, we first find the roots of the quadratic equation $x^2 - 18x + 68 = 0$. Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, with $a = 1$, $b = -18$, and $c = 68$, we find the roots to be $x = 9 \pm \sqrt{13}$. Since it's a parabola opening upwards (coefficient of $x^2$ is positive), $f(x) \geq 0$ outside the interval between the roots, hence the correct answer is $( - \infty , 9 - \sqrt { 13 } ] \cup [ 9 + \sqrt { 13 } , \infty )$.

Solve the Problem f(x)=x2−18x+68f ( x ) = x ^ { 2 } - 18 x + 68f(x)=x2−18x+68

Solve the Problem $f ( x ) = x ^ { 2 } - 18 x + 68$