The function g is defined as g(x)=ax^2+bx-6a, where a and b are positive constants. The function has a line of symmetry at x=-3. Which of the following could NOT be a coordinate pair on the graph of y=g(x)?

A) (0,-3)

B) (0,3)

C) (-3,0)

D) (3,3)

E) (-3,-3)

Solveit
Dec 10, 2015

#1**+25 **

Hi Solveit :))

The function g is defined as g(x)=ax^2+bx-6a, where a and b are positive constants. The function has a line of symmetry at x=-3. Which of the following could NOT be a coordinate pair on the graph of y=g(x)?

A) (0,-3)

B) (0,3)

C) (-3,0)

D) (3,3)

E) (-3,-3)

\(y=ax^2+bx-6a\)

\(\frac{-b}{2a}=-3\\ -b=2a*-3\\ -b=-6a\\ b=6a\\ \mbox{so the equation becomes}\\ y=ax^2+6ax-6a\\ y=a(x^2+6x-6) \)

If x=0 then

y=a(-6) <0 so (0,3) CANNOT be a point but (0,-3) could be

If x=-3

y=a(9-18-6) <0 so (-3,0) CANNOT be a point but (-3,-3) could be

If x=+3

y=a(9+18-6)>0 so (3,3) can be a point

SO

**(0,3) and (-3,0) cannot be points on the parabola.**

Melody
Dec 11, 2015

#1**+25 **

Best Answer

Hi Solveit :))

The function g is defined as g(x)=ax^2+bx-6a, where a and b are positive constants. The function has a line of symmetry at x=-3. Which of the following could NOT be a coordinate pair on the graph of y=g(x)?

A) (0,-3)

B) (0,3)

C) (-3,0)

D) (3,3)

E) (-3,-3)

\(y=ax^2+bx-6a\)

\(\frac{-b}{2a}=-3\\ -b=2a*-3\\ -b=-6a\\ b=6a\\ \mbox{so the equation becomes}\\ y=ax^2+6ax-6a\\ y=a(x^2+6x-6) \)

If x=0 then

y=a(-6) <0 so (0,3) CANNOT be a point but (0,-3) could be

If x=-3

y=a(9-18-6) <0 so (-3,0) CANNOT be a point but (-3,-3) could be

If x=+3

y=a(9+18-6)>0 so (3,3) can be a point

SO

**(0,3) and (-3,0) cannot be points on the parabola.**

Melody
Dec 11, 2015

#2**+5 **

Thanks melody for your answer ! :)

but there is no two answers, teacher said that parabola can be inverted and it means that it can be (-3,0) :D

Solveit
Dec 11, 2015

#3**+5 **

Thanks, Melody........I answered the same question here : https://web2.0calc.com/questions/little-help_1 ......and got the same result that you did....

Solveit -- If the parabola were inverted, "a" would have to be negative......but.......one of the conditions is that a,b > 0

Maybe I missed something???

CPhill
Dec 12, 2015

#4**0 **

Yes, it has to be concave up because "a" (the leading coefficient is positive)

Just like CPhill said :)

Melody
Dec 12, 2015

#5**+5 **

The function g is defined as g(x)=ax^2+bx-6a, where a and b are positive constants.

The function has a line of symmetry at x=-3.

hence -b/(2a)=-3 b=6a

I might try looking at it a different way

you think (-3,0) should be a possible point on the parabola.

g(-3)= a(-3)^2 +b*-3 -6a

g(-3)= 9a -3b -6a = 3a-3b = 3(a-b) = 3(a-6a) =3(-5a) = -15a <0 since a must be positive.

Therefore (-3,0) cannot be a point on the parabola.

Melody
Dec 12, 2015

#7**0 **

Hi Solveit,

**I am glad you are persisting. Persistence is how we learn difficult things. :)**

This was one of the constraints of your question

**The function has a line of symmetry at x=-3. **

Your graph is not correct because the axis of symmetry is x=0 not x=-3

g(x)=ax^2+bx-6a

for the axis to be x=-3

-b/2a must equal -3

-b/2a=-3

b=6a

so t he graph becomes

g(x)=ax^2+6ax-6a

**Here is the correct graph**

https://www.desmos.com/calculator/ws0izufvrt

NOW if/when (-3,0) is on the function then the function stops being a parabola and becomes a line.

Does a line have an axis of symmetry?

I have never heard anyone say so, but perhaps every perpendicular to a line is an axis of symmetry.

If every perpendicular to a line can be considered tro be an axis of symmetry then (-3,0) could be a point on the function.

SO

**I think that you are correct :)**

Melody
Jan 21, 2016