Solving x²-6x+8 > 0: Analyzing Quadratic Inequality with Graphs

The following functions are graphed below:

$f(x)=x^2-6x+8$

$g(x)=4x-17$

For which values of x is

$f(x)>0$ true?

To solve the inequality $f(x) > 0$, we first need to find the roots of the equation $f(x) = x^2 - 6x + 8$.

1. Find the roots of the quadratic equation:
The quadratic is $x^2 - 6x + 8$. This can be factored into:
$f(x) = (x - 2)(x - 4) = 0$.

2. Calculate the roots:
Setting each factor equal to zero gives the roots $x = 2$ and $x = 4$.

3. Determine the intervals defined by these roots:
The roots divide the x-axis into three intervals: $(-\infty, 2)$, $(2, 4)$, and $(4, \infty)$.

4. Test points in each interval to decide positivity:
- For $x < 2$, select $x = 1$: $f(1) = 1^2 - 6(1) + 8 = 3 > 0$. Thus, $f(x) > 0$ in $(-\infty, 2)$.
- For $2 < x < 4$, select $x = 3$: $f(3) = 3^2 - 6(3) + 8 = -1 < 0$. Thus, $f(x) < 0$ in $(2, 4)$.
- For $x > 4$, select $x = 5$: $f(5) = 5^2 - 6(5) + 8 = 3 > 0$. Thus, $f(x) > 0$ in $(4, \infty)$.

Therefore, the solution to $f(x) > 0$ is when $x < 2$ or $x > 4$.

The final solution is: $x < 2, 4 < x$.