Geometrical meaning of the zeroes of a polynomial.

Direct answer: Option A, 1/2 is a zero. A polynomial zero is any number r for which p(r)=0. On a graph, this means the point (r,0) lies on the x-axis. The graph touching the x-axis at x=1/2 therefore means p(1/2)=0, so 1/2 is a real zero. Crossing is not required: a graph may touch and turn back, as happens with an even-multiplicity root. Option A is correct. Option B is false because zeroes may be fractions, decimals, integers, or other real numbers. Option C confuses crossing with being a zero; touching also gives y=0. Option D changes 1/2 to its reciprocal 2 without any reason. The coordinate of contact is read directly from the x-axis. Exam cue: every meeting with the x-axis, whether crossing or touching, gives a zero.

A zero (root) of a polynomial is the x‑value where its graph meets the x‑axis — i.e. the x‑coordinate of the intercept. The given intercept \((-\sqrt{3},0)\) has x‑coordinate \(-\sqrt{3}\), so that is the zero. Option D is the full point (ordered pair), not the zero itself; option C is the y‑coordinate (0); option B has the wrong sign. Exam tip: from an x‑axis intercept (a,b) pick the first component a as the root.

A zero is related to the (x)-axis not a general (y)-intercept. A (y)-intercept gives a zero only if it is the origin.

Completing the square gives \(x^2+6x+10=(x+3)^2+1\). Since \((x+3)^2\ge0\), the minimum value is 1, so the parabola never reaches 0 and does not meet the x-axis. Equivalently, the discriminant is \(b^2-4ac=36-40=-4<0\), so there are no real roots. The closest distractor (touch once) is wrong because that requires discriminant zero. Exam tip: use the discriminant or complete the square to quickly decide if real x-intercepts exist.

The graph meets the x-axis where \(p(x)=0\). Solve \(2x^2-8=0\) to get \(x^2=4\), so \(x=\pm 2\). Equivalently factor as \(2(x-2)(x+2)\), giving zeros \(-2\) and \(2\). Choice B would follow from the wrong step \(x^2=16\); C is wrong because \(p(0)=-8\) (not zero); D is wrong because there are two zeros, both \(2\) and \(-2\), not only \(4\). Exam tip: set \(p(x)=0\) first and factor out common constants to simplify solving.

The point (9,0) lies on the x-axis, so for the polynomial p(x) we have \(p(9)=0\); hence x=9 is a root (zero). The fact that the graph goes from above to below shows the curve crosses the axis at that point, which indicates a root of odd order (often a simple root). Option B is wrong because y=9 is not implied by a point on the x-axis; option C is wrong because a nonzero constant polynomial does not cross the x-axis; option D contradicts the given point lying on the x-axis. Exam tip: A crossing of the x-axis implies a zero with odd multiplicity; touching-and-returning implies even multiplicity.

Points on the x-axis have y-coordinate zero, so set \(p(x)=0\): \(x^2-11x+30=0\). Factorising gives \(x^2-11x+30=(x-5)(x-6)\), so the roots are \(x=5\) and \(x=6\). Thus the graph meets the x-axis at (5,0) and (6,0). The closest distractor (B) has the signs reversed; option C lists y-axis intercepts, and option D confuses coefficients with roots. Exam tip: set the polynomial equal to zero and factorise (or use the quadratic formula) to find x-intercepts quickly.

Geometrically each (x)-axis intersection gives one real zero. A quadratic may have (0), (1), or (2) real zeroes.