If p(x) = x^2 - 9, at which points does its graph cut the x-axis?
Answer and explanation
Correct answer: (-3, 0) and (3, 0)
Set p(x)=0: \(x^2-9=0\). Factor: \((x-3)(x+3)=0\) so \(x=\pm3\). Zeros of the polynomial appear on the x-axis as points \((x,0)\), therefore \((-3,0)\) and \((3,0)\) are the intercepts. Options B and D list points on the y-axis (x=0), not x-axis intercepts; option C mistakes the roots as ±9 instead of ±3. Exam tip: either factor the quadratic or take square roots (\(x^2=9\) gives \(x=\pm3\)) to find intercepts quickly.
Frequently asked questions
What is the correct answer to this question?
(-3, 0) and (3, 0)
Why is this the correct answer?
Set p(x)=0: \(x^2-9=0\). Factor: \((x-3)(x+3)=0\) so \(x=\pm3\). Zeros of the polynomial appear on the x-axis as points \((x,0)\), therefore \((-3,0)\) and \((3,0)\) are the intercepts. Options B and D list points on the y-axis (x=0), not x-axis intercepts; option C mistakes the roots as ±9 instead of ±3. Exam tip: either factor the quadratic or take square roots (\(x^2=9\) gives \(x=\pm3\)) to find intercepts quickly.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Geometrical meaning of the zeroes of a polynomial..
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