Question
Graph of y = ax2 + bx + c = 0 is given adjacently. What conclusions can be drawn from this graph –
- a > 0
- b < 0
- c < 0
- All of the above
Hint:
So in this question, we have a graph given and we have to do it. As we can see the curve and it is of the parabola and by using the properties of the parabola and its equation, we can answer these questions easily.
The correct answer is: All of the above
As we can see from the graph we have a parabola curve and since it is opening in an upward direction. So we can say that a > 0 and
Hence, the option (a) is correct.
Here, we can see that the vertex of the parabola is located in the fourth quadrant , therefore it will be = 
On further solving this, we get
Therefore, the option (b) is also correct.
Since, at x=0 , the y intercept will be positive and from this, we can conclude that c < 0 and
Hence, the option (c) will also be correct
On checking all the options, and we can see all options are correct and
Therefore, we conclude that all the options available are correct.
Here we can see that the graph was given to us and us to take out the conclusion from that since we have the options available so I would suggest you to always start to check from the options because by the use of options we can see how easily we concluded this question.
Related Questions to study
For the quadratic polynomial f (x) = 4x2 – 8kx + k, the statements which hold good are
For the quadratic polynomial f (x) = 4x2 – 8kx + k, the statements which hold good are
The graph of the quadratic polynomial y = ax2 + bx + c is as shown in the figure. Then :
The graph of the quadratic polynomial y = ax2 + bx + c is as shown in the figure. Then :
The greatest possible number of points of intersections of 8 straight line and 4 circles is :
The students can make an error if they don’t know about the formula for calculating the number of points as mentioned in the hint which is as follows
The number point of intersection between two lines can be counted by finding the number of ways in which two lines can be selected out of the lot as two lines can intersect at most one point.
The number point of intersection between two circles can be counted by finding the number of ways in which two circles can be selected out of the lot multiplied by 2 as two circles can intersect at most two points.
The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.
The greatest possible number of points of intersections of 8 straight line and 4 circles is :
The students can make an error if they don’t know about the formula for calculating the number of points as mentioned in the hint which is as follows
The number point of intersection between two lines can be counted by finding the number of ways in which two lines can be selected out of the lot as two lines can intersect at most one point.
The number point of intersection between two circles can be counted by finding the number of ways in which two circles can be selected out of the lot multiplied by 2 as two circles can intersect at most two points.
The number point of intersection between two circles can be counted by finding the number of ways in which one circle and one line can be selected out of the lot multiplied by 2 as one circle and one line can intersect at most two points.
How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even position ?
Here we have obtained the total number of 9 digit numbers using the given digits. While finding the number of ways to arrange the odd digits in 5 even places, we have divided the 4! by 2! because the digit 3 were occurring two times and the digit 5 were occurring 2 times. Here we can make a mistake by conserving the number of even digits 4 and the number of odd digits 5, which will result in the wrong answer.
How many different nine digit numbers can be formed from the number 223355888 by rearranging its digits so that the odd digits occupy even position ?
Here we have obtained the total number of 9 digit numbers using the given digits. While finding the number of ways to arrange the odd digits in 5 even places, we have divided the 4! by 2! because the digit 3 were occurring two times and the digit 5 were occurring 2 times. Here we can make a mistake by conserving the number of even digits 4 and the number of odd digits 5, which will result in the wrong answer.
A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to :
A person predicts the outcome of 20 cricket matches of his home team. Each match can result either in a win, loss or tie for the home team. Total number of ways in which he can make the predictions so that exactly 10 predictions are correct, is equal to :
The foot of the perpendicular from the point 
on the line 
is
The foot of the perpendicular from the point on the line is
The point of intersection of the lines 
is
The point of intersection of the lines is
The line passing through 
and perpendicular to 
is
The line passing through and perpendicular to is
The equation of the line passing through 
is
The equation of the line passing through is
The length of the perpendicular from (-1, π/6) to the line 
is
The length of the perpendicular from (-1, π/6) to the line is
The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :
Alternatively, we can use the formula for the sum of numbers as
(n - 1)! 
(sum of digits) (11111 ............ntimes). We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.
The sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time is (repetition is not allowed) :
Alternatively, we can use the formula for the sum of numbers as
(n - 1)! (sum of digits) (11111 ............ntimes). We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.
Total number of divisors of 480, that are of the form 4n + 2, n 
0, is equal to :
We can also solve this question by writing 4n + 2 = 2(2n + 1) where 2n + 1 is always an odd number. So, when all odd divisors will be multiplied by 2, we will get the divisors that we require. Hence, we can say a number of divisors of 4n + 2 form is the same as the number of odd divisors for 480.
Total number of divisors of 480, that are of the form 4n + 2, n 0, is equal to :
We can also solve this question by writing 4n + 2 = 2(2n + 1) where 2n + 1 is always an odd number. So, when all odd divisors will be multiplied by 2, we will get the divisors that we require. Hence, we can say a number of divisors of 4n + 2 form is the same as the number of odd divisors for 480.
. 
. 15
have a common factor then 'a' is equal to
