## Algebra and Trigonometry 10th Edition

a) See graph b) $x=-2.1213,x=2.1213$
We are given the function: $f(x)=\dfrac{2x^2-9}{3-x}$ a) Graph the function. From the graph the zeroes are: $x\approx -2.121,x\approx 2.121$ b) Determine the zeros algebraically: $f(x)=0$ $\dfrac{2x^2-9}{3-x}=0$ $2x^2-9=0$ $x^2=\dfrac{9}{2}$ $x=\pm\dfrac{3}{\sqrt 2}=\pm 2.1213$