## Algebra and Trigonometry 10th Edition

Zeros of $f(x)$: $x=2,7$
$f(x)=\frac{x^2-9x+14}{4x}$ The zeros of a function are the x-values for when y-values equal zero. In other words, zeros are when $y=f(x)=0$. Since we are solving for the zeros of $f(x)$, set the function $f(x)=\frac{x^2-9x+14}{4x}$ equal to zero. We get the following equation: $\frac{x^2-9x+14}{4x}=0$ First, multiply both sides of the equation by $4x$ to simplify the denominator: $\frac{x^2-9x+14}{4x}(4x)=0(4x)$ $x^2-9x+14=0$ Now factor the left side of the equation by finding two numbers that multiply to get 14 and add to get -9. These two numbers are -2 and -7. So the factored left side looks as follows: $(x-2)(x-7)=0$ Next, you can see that the above equation holds true when $x-2=0$ or $x-7=0$. Solve each of those two equations individually. $x-2=0$ Add 2 to both sides. $x-2+2=0+2$ $x=2$ Now for the other equation: $x-7=0$ Add 7 to both sides. $x-7+7=0+7$ $x=7$ So, the zeros of $f(x)$ are $x=2,7$