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)
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.
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.
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
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???
Yes, it has to be concave up because "a" (the leading coefficient is positive)
Just like CPhill said :)
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.
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 :)