Contents
Question 1
Examine, seeing the graph of the polynomials given below, whether they are a linear or quadratic polynomial or neither linear nor quadratic polynomial:
(i)
Sol :
(i) In general, we know that for a linear polynomial ax + b, a≠0, the graph of y = ax + b is a straight line which intersects the x – axis at exactly one point.
And here, we can see that the graph of y = p(x) is a straight line and intersects the x – axis at exactly one point. Therefore, the given graph is of a Linear Polynomial.
(ii)
Sol :
(ii) Here, the graph of y = p(x) is a straight line and parallel to the x – axis . Therefore, the given graph is of a Linear Polynomial.
(iii)
Sol :
(iii) For any quadratic polynomial ax2 + bx + c, a ≠ 0, the graph of the corresponding equation y = a x2 + bx + c has one of the two shapes either open upwards likeor open downwards likedepending on whether a > 0 or a < 0. (These curves are called parabolas.)
Here, we can see that the shape of the graph is a parabola. Therefore, the given graph is of a Quadratic Polynomial.
(iv)
Sol :
(iv) For any quadratic polynomial ax2 + bx + c, a ≠ 0, the graph of the corresponding equation y = ax2 + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a > 0
or a < 0. (These curves are called parabolas.)
Here, we can see that the shape of the graph is parabola. Therefore, the given graph is of a Quadratic Polynomial.
(v)
Sol :
(v) The shape of the graph is neither a straight line nor a parabola. So, the graph is not of a linear polynomial nor a quadratic polynomial.
(vi)
Sol :
(vi) The given graph have a straight line but it doesn’t intersect at x – axis and the shape of the graph is also not a parabola. So, it is not a graph of a quadratic polynomial. Therefore, it is not a graph of linear polynomial or quadratic polynomial.
(vii)
Sol :
(vii) The shape of the graph is neither a straight line nor a parabola. So, the graph is not of a linear polynomial or a quadratic polynomial.
(viii)
Sol :
(viii) The shape of the graph is neither a straight line nor a parabola. So, the graph is not of a linear polynomial or a quadratic polynomial.
Question 2
The graphs of y – p(x) are given in the figures below, where p(x) is a polynomial. Find the number of zeros in each case.
(i)
Sol :
(i) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.
(ii)
Sol :
(ii) Here, the graph of y = p(x) intersects the x – axis at three points. So, the number of zeroes is 3.
(iii)
Sol :
(iii) Here, the graph of y = p(x) intersects the x – axis at one point only. So, the number of zeroes is 1.
(iv)
Sol :
(iv) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1.
(v)
Sol :
(v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.
(vi)
Sol :
(vi) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1.
Question 3
The graphs of y = p(x) are given in the figures below, where p(x) is a polynomial Find the number of zeroes in each case.
(i)
Sol :
(i) Here, the graph of y = p(x) intersect the x – axis at zero points. So, the number of zeroes is 0.
(ii)
Sol :
(ii) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.
(iii)
Sol :
(iii) Here, the graph of y = p(x) intersects the x – axis at four points. So, the number of zeroes is 4.
(iv)
Sol :
(iv) Here, the graph of y = p(x) does not intersects the x – axis. So, the number of zeroes is 0.
(v)
Sol :
(v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.
(vi)
Sol :
(vi) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.