Square Function

Question

Let F be a quadratic function given by F(x)=ax2+bx-2. If x=3 is the axis of symmetry of the graph of F and F(3)=>0, then it is fulfilled with certainty:

Which is the correct answer.

A) a>0, b>0

B) a>0, b<0

C) a<0, b>0

D) a<0, b<0

Please explain your answer

Really appreciate your help :)

Answer 1

You can rule out A and D, leaving B and C as the only possible correct answers

x=3 is the line of symmetry

Look at the first part of the quadratic formula x= -b/2a + or - (1/2a)sqr(b^2-4ac),

the -b/2a part = +3

x=-b/2a = 3. That's only true, if a and b have opposite signs. That narrows it down to either B or C

F(3) > or = 0 means the vertex is above the x axis and the parabola is downward opening, given that the y intercept = -2 = c. That means a<0

all downward opening parabolas have a, the coefficient of the x^2 term, less than 0.

C is correct. a<0 b>0

Answer 2

The x-value of the vertex (which will be on the axis of symmetry) can be found using the equation x = -b/(2a). Since we are told that the axis of symmetry is x = 3 then we can set -b/(2a) equal to 3:

-b/(2a) = 3

-b = 6a

b = -6a

So we can now write the equation as:

f(x) = ax² - 6ax - 2

Since we know that f(3) > 0 we can say:

a(3)² - 6a(3) - 2 > 0

9a - 18a - 2 > 0

-9a > 2

a < -2/9

Using our original b = -6a, we can say that a = -b/6 and, plugging that into a < -2/9 we get:

-b/6 < -2/9

-b < -12/9

b > 12/9

b > 4/3

So "a" is less than zero and "b" is greater than zero