Geometrical meaning of the zeroes of a polynomial.

Here \(p(x)\) denotes the polynomial's value (the y‑coordinate). \(p(-1)=0\) means at \(x=-1\) the value is \(y=0\), so the graph passes through the point \((-1,0)\). Option B \((1,0)\) would be correct only if \(p(1)=0\) — it's a sign‑error distractor. Options C and D correspond to values at \(x=0\) and are irrelevant unless \(p(0)=-1\) or \(p(0)=1\) were given. Exam tip: whenever \(p(a)=0\), the x‑coordinate of the zero is \(a\) and the point is \((a,0)\).

\(p(6)=0\) means the function value is zero at \(x=6\), so the graph meets the x-axis (where \(y=0\)) at the point \((6,0)\). The closest distractor \((-6,0)\) is a sign error — nothing in the given information implies \(p(-6)=0\). Exam tip: whenever you are given \(p(a)=0\), the corresponding x-intercept is always \((a,0)\).

Zeroes are the (x)-coordinates of the points where the graph cuts the (x)-axis. So the zeroes are (2) and (7).

When a parabola touches the (x)-axis at its vertex, it has one real zero. Its (x)-coordinate is (4).

If the zeroes are (-3) and (2), the graph cuts the (x)-axis at those (x)-values. So the points are ((-3,0)) and ((2,0)).

Zeroes are found from intersections with the (x)-axis, not the (y)-axis. So if there is no (x)-axis intersection, there are (0) real zeroes.

The governing concept is the graphical meaning of a zero of a polynomial. A number r is called a zero of p(x) when p(r) = 0. On the graph y = p(x), the point corresponding to x = r therefore has coordinates (r, p(r)) = (r, 0). Hence its vertical coordinate, or y-value, is zero. Geometrically, every real zero is represented by a point where the polynomial graph meets or touches the x-axis. The options 1 and -1 are possible y-values at other points, but not at a zero. The value cannot be undetermined because the defining condition of a zero itself fixes p(r) as 0.

A point on the x-axis has \(y=0\). So set \(2x-6=0\), giving \(2x=6\) and \(x=3\). Hence the zero is 3. The option -3 arises from a sign or arithmetic mistake; 6 or -6 are values of y (the y-intercept is -6), not the x-intercept. Exam tip: to find the zero of a linear polynomial from its line, always put \(y=0\) and solve for x.

The x-axis corresponds to y = 0, so set y = 0 in \(y = x + 7\): \(0 = x + 7\) ⇒ \(x = -7\). Thus the zero (x-intercept) is -7. Option A (7) is the common sign-error distractor; options C and D do not satisfy the equation. Exam tip: to find an x-intercept of any line, put y = 0 and solve for x.

If the graph touches the x-axis at (5,0) then the polynomial evaluates to zero at x=5, i.e. \(p(5)=0\). Hence x=5 is a zero. "Touching" typically implies the root has even multiplicity, but the essential fact here is the value is zero. Options B and D are incorrect because the contact point has x-coordinate 5, not 0 or -5; option C is wrong because touching the axis means the polynomial is zero at that point. Exam tip: use the x-coordinate of the contact point and substitute into \(p(x)\) to verify a zero; if it only touches (doesn't cross), expect even multiplicity.

Each distinct (x)-axis intersection represents one real zero. Therefore (4) intersections give (4) real zeroes.

By definition a number a is a zero of a polynomial iff \(p(a)=0\). Here \(p(2)=5\), which is not zero, so 2 is not a zero. The degree of the polynomial (linear, quadratic, etc.) does not affect this: if \(p(a)\neq0\) it cannot be a root. Exam tip: To check whether a value is a root, evaluate \(p(a)\) directly — roots give result 0; remainders on dividing by \((x-a)\) equal \(p(a)\).

For a polynomial function the graph contains points of the form (x,p(x)). If p(a)=0 then the point (a,p(a)) becomes (a,0), so the graph must pass through (a,0). The closest distractor, (-a,0), is also on the x-axis but would be correct only if p(-a)=0. Exam tip: Always express a point as (x,p(x)) and substitute the given x to find the corresponding point on the graph.

Zeroes are the (x)-values where the graph cuts the (x)-axis. So the zeroes are (1), (3), and (6).

Points where a parabola meets the x-axis are its real zeros. Two distinct intersections therefore mean two distinct real zeros; here the zeros are x = -1 and x = 5. Option B (one) would correspond to a tangent (a repeated root), option C (none) is impossible because intersections are given, and option D (three) is impossible for a quadratic since a parabola (degree 2) can have at most two real zeros. Exam tip: Count x-axis intersections for real zeros; for quadratics the maximum is 2 and a single intersection implies a repeated root.

Touching the x-axis means the parabola's vertex lies on the axis and the quadratic has a repeated root — one distinct real zero (a double root). In terms of discriminant, this corresponds to \\(\Delta=0\\). So there is exactly one real zero: x=2. Why other options are wrong: 'Two' would require the parabola to intersect the axis at two distinct points; 'None' applies when the parabola does not meet the axis at all; 'Infinite' is impossible for a polynomial of finite degree. Exam tip: for a double root of a quadratic check that \\(\Delta=0\\).

Real zeroes of a quadratic polynomial are found from points where it meets the (x)-axis. Two distinct intersections give two real zeroes.

For real zeroes, the graph must meet the (x)-axis. If it does not meet the (x)-axis, there is no real zero.

The number of distinct times the graph meets the (x)-axis gives the number of real zeroes. Check this first while reading a graph.

An x-intercept of (-4, 0) means the polynomial evaluates to 0 at x = -4, so x = -4 is a root. Option B (4) is incorrect because it has the opposite sign; options C (0) and D (1) would only be correct if those x-values were the intercepts, which they are not. Exam tip: the x-coordinate of an x-intercept gives the root directly; if unsure, substitute the value into the polynomial to verify.