Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter P - P.1 - Graphs and Models - Exercises - Page 9: 79

Answer

This statement is true.

Work Step by Step

The function $y=ax^2+bx+c$ is a quadratic polynomial function. The intercepts with the $x$ axis will occur when $y=0$. Putting $y=0$ we get $$ax^2+bx+c=0$$ and the solutions for $x$ determine the intercepts with the $x$ axis. Now this equation is a quadratic equation and its' solutions are $$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ Since $b^2-4ac\gt0$ its root will exist and it will be bigger that zero so there will be two different solutions of the equation: $$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a};\quad x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$ which means we will have two intercepts, one in $x_1$ and the other in $x_2$.
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