If the diagram in Fig. 2.22 shows the graph of the polynomial f(x) = ax² + bx + c, then
(a) a > 0, b < 0 and c > 0
(b) a < 0, b < 0 and c < 0
(c) a < 0, b > 0 and c > 0
(d) a < 0, b > 0 and c < 0
Answers
SOLUTION :
Option (a) is correct : (a) a > 0, b < 0 and c > 0
From the figure, the graph of polynomial p(x) is a parabola open upwards, therefore a >0.
y = ax²+bx+c cuts Y- axis at P. On Putting x = 0 in y = ax²+bx+c , We get y = c. Hence , the coordinates of P is (0,y) or (0,c).It is clear that P lies on OY. Therefore c > 0.
**For any quadratic polynomial ax² + bx + c , the zeros are precisely the x- coordinates of the points where the graph of y = ax² + bx + c intersects the X- axis.
**For any quadratic polynomial the graph of the corresponding equation y = ax² + bx + c has one of the two shapes which are known as parabola either open upwards or open downwards. If a > 0 then the shape of parabola is open upwards or a< 0 then the shape of parabola is open downwards.
•If the graph intersects the X-axis AT TWO POINTS then a quadratic polynomial HAS TWO DISTINCT ZEROS. D= b² - 4ac > 0.
•If the graph intersects or touches the X-axis at EXACTLY ONE POINT then a quadratic polynomial has TWO EQUAL ZEROES (ONE ZERO).D= b² - 4ac = 0.
•If the graph is either completely above X-axis or completely below X-axis axis i.e it DOES NOT INTERSECT X-AXIS axis at any point .Then the quadratic polynomial HAS NO ZERO .D= b² - 4ac < 0.
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Answer:
Step-by-step explanation:
Option (a) is correct : (a) a > 0, b < 0 and c > 0
From the figure, the graph of polynomial p(x) is a parabola open upwards, therefore a >0.
y = ax²+bx+c cuts Y- axis at P. On Putting x = 0 in y = ax²+bx+c , We get y = c. Hence , the coordinates of P is (0,y) or (0,c).It is clear that P lies on OY. Therefore c > 0.