If \(p(x)=x^2-16\), what are the x-axis intersections (x-intercepts) of its graph?
Answer and explanation
Correct answer: (4,0) and (-4,0)
X-intercepts occur where the polynomial equals zero, i.e. set \(p(x)=0\). Solving \(x^2-16=0\) gives \(x^2=16\) so \(x=\pm4\). Therefore the x-intercepts are \((4,0)\) and \((-4,0)\). Option A lists y-axis-like points that are not zeros here; options C and D are incorrect because substituting those x-values does not make \(p(x)=0\) (for example \(p(16)=256-16\neq0\)). Exam tip: always solve \(p(x)=0\) to find zeros and report them as \((x,0)\).
Frequently asked questions
What is the correct answer to this question?
(4,0) and (-4,0)
Why is this the correct answer?
X-intercepts occur where the polynomial equals zero, i.e. set \(p(x)=0\). Solving \(x^2-16=0\) gives \(x^2=16\) so \(x=\pm4\). Therefore the x-intercepts are \((4,0)\) and \((-4,0)\). Option A lists y-axis-like points that are not zeros here; options C and D are incorrect because substituting those x-values does not make \(p(x)=0\) (for example \(p(16)=256-16\neq0\)). Exam tip: always solve \(p(x)=0\) to find zeros and report them as \((x,0)\).
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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