1) you can find two different solutions for this because you can change a, b and c. You could factor like this:
(x+6)(x+2) and a = 1 an b = 8. This will give you one parabola. When x = 0 then y = 12

Or you can factor like this:
(2x+4)(x+6) and a = 2 and b = 20. This will give you another parabola. When x = 0 then y = 24

Insufficient

2) This alone will not solve as we can have a variety of different equations by changing b and c.

Together, we can have the following factored equations:

(2x+4)(x+6)
(x+2)(2x+12)

Both of these give a y-intercept of 24

Answer = C

karthikpandian19 wrote:If f(x) = ax^2 + bx + c, with "a" not equal to zero, at what point does the graph of the function f(x) intersect the y-axis?

1. The graph of f(x) intersects the x-axis exactly twice, at (-6,0) and (-2,0).
2. a=2

Three ways of doing this:

a. Way 1: Reasoning using linear equations:

We have y=ax^2+bx+c as the quadratic function. We need to find c. Then,
1. The curve passes through (-6,0) and (-2,0). This will give us two independent equations. We need three to be able to solve for three variables. Insufficient.

2. a=2 gives us a single linear equation. Insufficient.

We can see that together these three will have system of 3 equations with 3 unknowns. Sufficient to solve for c. Hence C.

b. Way 2: Reasoning using properties of quadratics. We need to find c.

1. We are told where the quadratic intercepts the x-axis. We don't even know whether the quadratic is open upwards or downwards (whether a<0, or a>0). Insufficient.

2. a=2 only tells us that parabola is open upwards. So we can deduce that c (the y-intercept) has a negative value. Insufficient.

Together, we know that c= -x^2-bx when y=0. Since we know two values of x, we can determine c. Sufficient. Hence C.

c. Way 3: Algebraic, minimum reasoning. We are looking for c.

1. The curve passes through (-6,0) and (-2,0)
= > a(-6)^2+b(-6)+c = 0 and a(-2)^2+b(-2) + c =0
= > 36a-6b+c = 0 and 4a-2b+c=0
(multiplying second equation by 3 and subtracting the two equations we get)
= > 36a-6b+c - 12a - 6b -3c = 0
= > 24a = 2c => c=12a.
c depends on a, Insufficient.

2. a=2. We don't know what c is. Insufficient.

Together c = 12a and a=2 => c=24. Sufficient. Hence C.

OA is C

eagleeye wrote:

karthikpandian19 wrote:If f(x) = ax^2 + bx + c, with "a" not equal to zero, at what point does the graph of the function f(x) intersect the y-axis?

1. The graph of f(x) intersects the x-axis exactly twice, at (-6,0) and (-2,0).
2. a=2

Three ways of doing this:

a. Way 1: Reasoning using linear equations:

We have y=ax^2+bx+c as the quadratic function. We need to find c. Then,
1. The curve passes through (-6,0) and (-2,0). This will give us two independent equations. We need three to be able to solve for three variables. Insufficient.

2. a=2 gives us a single linear equation. Insufficient.

We can see that together these three will have system of 3 equations with 3 unknowns. Sufficient to solve for c. Hence C.

b. Way 2: Reasoning using properties of quadratics. We need to find c.

1. We are told where the quadratic intercepts the x-axis. We don't even know whether the quadratic is open upwards or downwards (whether a<0, or a>0). Insufficient.

2. a=2 only tells us that parabola is open upwards. So we can deduce that c (the y-intercept) has a negative value. Insufficient.

Together, we know that c= -x^2-bx when y=0. Since we know two values of x, we can determine c. Sufficient. Hence C.

c. Way 3: Algebraic, minimum reasoning. We are looking for c.

1. The curve passes through (-6,0) and (-2,0)
= > a(-6)^2+b(-6)+c = 0 and a(-2)^2+b(-2) + c =0
= > 36a-6b+c = 0 and 4a-2b+c=0
(multiplying second equation by 3 and subtracting the two equations we get)
= > 36a-6b+c - 12a - 6b -3c = 0
= > 24a = 2c => c=12a.
c depends on a, Insufficient.

2. a=2. We don't know what c is. Insufficient.

Together c = 12a and a=2 => c=24. Sufficient. Hence C.