Chapter 5: Exercises with Solutions
Chapter 5: Exercises with Solutions
Chapter 5: Exercises with Solutions
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Section 5.3Zeros of the Quadratic45. To find the zeroes, set f(x) = 0 and factor.0 = −4x 2 + 4x + 150 = −(4x 2 − 4x − 15)0 = −(2x + 3)(2x − 5)By the zero product property, either2x + 3 = 0 or 2x − 5 = 0.Solve these linear equations independently.x = −3/2 or x = 5/247. To find the zeroes <strong>with</strong> your calculator, press 2nd TRACE to access the CALCmenu and choose 2:zero.Use the left arrowto move the cursoralong the curve untilit is to the left of thefirst zero. Hit ENTER.Use the right arrow tomove the cursor until itis to the right of thesame zero. Hit ENTER.Finally hit ENTERnear that same zerofor the guess, andyou get the zero.Repeat this process for the second zero. We get (3.5, 0) and (−5, 0).50yf(x)=2x 2 +3x−35−10(−5,0) (3.5,0)x10−50To find the zeroes algebraically, set y = 0 and solve for x:Version: Fall 2007