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# little help :)

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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)

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.

Dec 11, 2015
edited by Melody  Dec 11, 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
edited by Melody  Dec 11, 2015
#2
+5

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

Dec 11, 2015
#3
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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???   Dec 12, 2015
#4
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Yes, it has to be concave up because "a"  (the leading coefficient is positive)

Just like CPhill said :)

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.

Dec 12, 2015
#6
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Here is the graph that (-3,0) can be:

https://www.desmos.com/calculator/rxe34pziyx

Jan 21, 2016
edited by Solveit  Jan 21, 2016
edited by Solveit  Jan 21, 2016
#7
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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  :)

Jan 21, 2016
#8
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Jan 21, 2016
edited by Solveit  Jan 21, 2016
#9
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yes i understood everything thanks Melody ! :)

Jan 21, 2016