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Exercise 22

Chapter P, Section P.1, Page 8
Calculus 9th Edition by Bruce H. Edwards, Ron Larson
ISBN: 9780547167022

Solution

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By definition:

To find x-intercept of a graph , let y be zero and solve the equation for x.

y2=x34x02=x34x0=x34x0=x(x24)0=x(x+2)(x2)\begin{align*} y^{2}&= x^{3}-4\cdot x\\ 0^{2}&=x^{3}-4\cdot x \\ 0&=x^{3}-4\cdot x \\ 0&=x\cdot \left(x^{2} -4\right) \\ 0&=x\cdot \left(x +2\right) \cdot \left(x-2 \right) \\ \end{align*}

Then, x-intercept are x=0x=0 ,x=2x=-2 ,x=2x= 2.

And to find y-intercept of a graph , let x be zero and solve the equation for y.

y2=x34xy2=0340y2=0y=0\begin{align*} y^{2}&= x^{3}-4\cdot x\\ y^{2}&= 0^{3}-4\cdot 0 \\ y^{2}&=0 \\ y&=0\\ \end{align*}

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