find a quadratic expression for g( x )

[bumblebee123]

Can anyone give me a nudge in the right direction for this question- I'm not sure where to start.

question: the curve with the equation y = 7 - 2x - x^2 has a single stationary point whose coordinates are ( -1, 8 )

a vertical translation applied to the curve to give curve y = g (x). The stationary point of the curve y + g ( x ) lies on the x-axis

a) find a quadratic expression for g( x )

a horizontal translation is applied to the curve y = 7 - 2x - x^2 to give the curve y = h (x)

the stationary point of the curve y = h (x) lies on the y-axis

b) find a quadratic expression for h (x)

 

[Dr.Peterson]

The problem could be stated more clearly. They are asking (a) what vertical translation will move (-1, 8) to the x-axis, and (b) what horizontal translation will move (-1, 8) to the y-axis. Then apply those translations to the given function, f(x) = 7 - 2x - x^2 , to obtain g(x) and h(x).

 

[bumblebee123]

for part a) would this mean y = 0

 

[MarkFL]

I think I would begin by writing the given function in vertex form \(y=a(a-h)^2+k\) where the vertex is at \((h,k)\):

[MATH]f(x)=7-2x-x^2=-(x^2+2x)+7=-(x^2+2x+1)+7+1=-(x-(-1))^2+8[/MATH]
We can see now that the vertex (which is the turning point for a quadratic function) of \(f(x)\) is at \((-1,8)\) as stated.

Now, it's just a matter of applying the required translations to move the vertex to the desired locations.

 

[Dr.Peterson]

for part a) would this mean y = 0

Yes, you want to move (-1, 8) down to (-1, 0), with y=0. What transformation does that? What will that do to the equation?

 

[Steven G]

for part a) would this mean y = 0

You want to take away 8 from each and every y value so that (-1,8) will go to (-1,0). How do you do that? Also graph any curve and move each and every point down 8 units just to see what is going on.

 

[bumblebee123]

to move every y value 8 less than the function, would it be y = f(x) - 8

 

[bumblebee123]

so g ( x) = -1 - 2x - x^2

 

[Dr.Peterson]

Yes, and you can verify that g(-1) = 0, and that this is its stationary point.

Now do the second part.

 

[bumblebee123]

I have no idea what to do for part b)

 

[Dr.Peterson]

Go back and read post #2. How do you do a horizontal translation?

 

[bumblebee123]

(b) what horizontal translation will move (-1, 8) to the y-axis.

y = f ( x -1 )

 

[Steven G]

(b) what horizontal translation will move (-1, 8) to the y-axis.

y = f ( x -1 )

Yes.

 

[bumblebee123]

how do I make that into a quadratic expression?

 

[Deleted member 4993]

how do I make that into a quadratic expression?

If f(m) = 7 - 2m - m^2

what expression will define f(n)?

What expression will define f[(n-1)]?

 

[bumblebee123]

7 - 2x - x^2 - 1

 

[Deleted member 4993]

7 - 2x - x^2 - 1....................... Incorrect

f[(x-1)]

= 7 - 2(x-1) - (x-1)^2

= 7 -2x +2 - (x^2 -2x +1)

= 7 - 2x + 2 - x^2 + 2x -1

= 8 - x^2

 

[bumblebee123]

I'm not sure what's being done?

 

[Deleted member 4993]

how do I make that into a quadratic expression?

I am responding to your question and your follow-up response.

Exactly where did you get lost?

 

[bumblebee123]

well, the original equation is
y = 7 - 2x - x^2
Ididn't know how to turn that into y = f ( x -1 )

looking back over what you did, did you just replace x in the original equation, with (x-1)? Because after looking over it several times, it makes sense
y = f ( x -1 ) = 7 - 2(x-1) - (x-1)^2