फलन \(y=x^2-4\) का ग्राफ (x)-अक्ष को किन बिंदुओं पर काटता है?
At which points does the graph of \(y=x^2-4\) cut the (x)-axis?
Explanation opens after your attempt
A. ( \left\(-2,0\right\),\left\(2,0\right\) )
Concept
Set (y=0) on the (x)-axis. From \(x^2-4=0\), \(x=\pm 2\).
Why this answer is correct
The correct answer is A. ( \left\(-2,0\right\),\left\(2,0\right\) ). Set (y=0) on the (x)-axis. From \(x^2-4=0\), \(x=\pm 2\).
Exam Tip
(x)-अक्ष पर (y=0) रखें। \(x^2-4=0\) से \(x=\pm 2\) मिलता है।
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