Shape Of Parabola

putting x = 0 then we can see from graph that c is positive, also both roots are positive, then the equation can be written as, (x-d)(x-e) = x^2 - (d+e)x + de where d and e are its roots. on comparing we find that b is negative. therefore bc<0.

C is positive and b is negative...

If we take the derivative ${ dy }/{ dx } = 2x+b$ . Then when x=0, ${ dy }/{ dx }$ must be negative (we can see negative slope there). So, b is negative. We also know that when x=0, y>0, then c >0. :) :)

The y-intercept in the equation y = ax^2 + bx + c is "c". (Just set the value of x equal to zero and solve for "y".)

And in the graph, c is in the positive y-axis. Therefore, "c" is positive. Or that "c > 0".

Find the vertex of the parabola. Transform the equation in standard form: y = a(x-h)^2 + k where the vertex is at point (h, k).

y = x^2 + bx + c
y = (x^2 + bx) + c
y = [x^2 + bx + (b^2/4)] + c - (b^2/4)
y = [x + (b/2)]^2 + c - (b^2/4)

y = [x - (-b/2)]^2 + (4c-b^2)/4

The vertex is at (-b/2, (4c-b^2)/4).

Observe that the vertex in the graph is located at the fourth quadrant. That makes the x-coordinate a positive and the y-coordinate a negative.

So, "-b/2" must be positive. And it's true if "b is negative". Or we write "b < 0".

Together, b<0 and c>0, we know for sure that "bc" will always be less than zero. That is "bc < 0".

Roots are positive hence their sum and product must be positive. So using vieta's, we have -b>0 hence b<0 and c>0 hence bc<0.

put x=0,then y=c and from the graph y>0 so c>0. now look at the roots of the equation,both are positive, so sum of roots>0 so -b/a>0 now parabola opens upwards so a>0 hence,b<0 and c>0 so bc<0

Someone has very intelligently put the 4th option [b^2 - 9c > 0], now that implies D > 5c, notice people, that D is not just the discriminant but in this case it is also the square of the distance between the roots, If we look closely, sqrt(D) = distance between the roots which is atleast approx. 3 times the height of the parabola at the origin. Root(D) > 3c, then D > 9c, definitely D>5c, that is b^2 - 4c >5c, or that b^2 - 9c>0 I thought it was very beautiful, but when I realized option c is true aswell, I was disappointed. 4th option with respect to the figure is absolutely true, but too bad the question wasnt "this-brilliant" :P

f(0) = c >0 and sum of roots = +ve => -b > 0 => b<0 so bc<0

for a positive value of x , y is going to be zero assume the value or the root is p(+ive) then p^2+b x p + c=0 only one condition satisfies that b and c should be of different signs. i.e. bc<0

Since from the graph we can observe that the quad. has 2 positive real solutions => the sum of roots is +ive , product of roots is also +ive => b<0 : c>0 => bc <0.

when we put x=0 then y is positive(from figure so c>0 and for some values of x y is negative as c is positive and x^2 is also positive so b must be negative so answer is bc<0

Since the parabolla cut Ox in 2 positive x value, the equation f(x) = 0 has 2 positive roots.

Therefore, S > 0 and P> 0 => b<0 and c > 0 => bc<0

x1 > 0 and x2 > 0 b = -(x1+x2) so b < 0 c = x1 * x2 so c >0 therefore bc < 0

if we will put x=0, in function y, then it is clear that y=c>0.....that means c is positive.. and by equating dy/dx=0,we get the value of x= -b/2(value of x at which y is minimum).....which is greater than zero from graph so -b/2>0 => b<0 ... b is negative.. ANSWER= bc<0.

at x=0 we can see that graph is positive=> c>0, similarly derivative is zero at x=-b/2, which from graph turn out to be positive so b might be negative. Hence bc should be less than 0.

Well, obviously $f(x) = (\alpha x - \beta)^{2} - \gamma = (\alpha x)^{2} - 2 \alpha \beta x + \beta^{2} - \gamma = ax^{2} + bx + c$ with $\alpha \geq 0$ , $\beta \geq 0$ and $\gamma \geq 0$ . Then $a = \alpha^{2} \geq 0$ , $b = -2\alpha \beta \leq 0$ and $c = \beta^{2} + \gamma \geq 0$ . Then notice that

$bc = -2 \alpha \beta^{3} - 2 \alpha \beta \gamma = -2 \alpha \beta (\beta^{2} + \gamma) \leq 0$

First note that x^2+bx+c=0 has two distinct solutions. Which implies b^2-4c>=0. two solutions are x= -(b/2)+/- Sqrt(b^2-4c)/2. And the minima of the graph is at -b/2. But its obvious that minima happens at positive x. So b<0. Now the solution -(b/2)-Sqrt(b^2-4c)>0. This straight away gives us c>0. together bc<0.

Since both roots are positive hence sum of roots (that is c) and product of roots(that is '-b') are positive. Hence c is positive and b is negative hence b*c is negative

since both the root of the equation are positive therefore, sum of roots must be positive thus -b>0 therefore,b<0 product of roots positive thus c>0 therefore bc<0

1)sum of roots = -b/a which is definitely positive as both roots are positive, 2)product of roots=c/a which is again positive as both roots are positive 3)as its open upwards parabola thus a is also positive so c is also positive by statement 2 and b is negative by statement 1. thus bc<0

Clear from the graph are the entities: c >0 , b<0 hence the answer.

it's a equation of a parabola whose vertex at (-b/2,c),and here in the picture the vertex of the parabola situated on the 4th quadrant where the co-ordinate of the points are of the form(+,-).So it is very clear that b<0 and c>0,i.e. b*c<0

When x is 0, we will have Y as the value of c. 0^2 + b*0 + c = c

So as the graph shows, c > 0

The critic point ( the point of minimum value of y) is given by -b/2a at the x axis in every quadratic function. We know by the graph that it happens when x > 0. So...

-b/2a > 0 ( a = 1)

-b > 0

b < 0

So b is negative and c positive. So bc is negative ( < 0)

:)

From the fact that the parabola crosses only at the positive x-axis, the roots are positive numbers, $m$ and $n$ which make the factored form of the equation of the parabola $y = (x - m)(x - n)$ which leads to $y = x^2 - (m + n)x + mn$ .

$b$ is a negative number (from $-(m + n)x = bx \Rightarrow -(m + n) = b$ ) and $c$ is a positive number (from $mn = c$ ) and thus the product $bc$ is a negative number.

the lowest point is at y= -D/4a.. Here, a=1, and -D = 4c-b^2, so this needs to be less than 0, this means that b^2-4c needs to be greater than 0, but we have b^2-9c, about which we can't say anything.. So, its bc less than 0, as c comes out to be more than 0, and b to be less than 0..