Match the parabolas represented by the equations with their vertices.

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07-04-2018
Mathematics

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Match the parabolas represented by the equations with their vertices.

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mathmate mathmate · Answer 1 · 2018-04-07T22:37:34+02:00

Question asks us to match the list of parabolas, given the vertices.

For any given parabola
f(x)=ax^2+bx+c,the vertex is at x=-b/(2a), and y=-(b^2-4ac)/4a.

This means that given the vertex x=-b/(2a), we just have to compare with -b/2a of each parabola.
For
f1(x)=x^2+6x+8, -b/(2a)=-6/2=-3, so (-3,-1) is a candidate for the vertex.
f2(x)=2x^2+16x+28, -b/(2a)=-16/4=-4, so (-4,-4) is a candidate.
f3(x)=-x^2+5x+14, -b/(2a)=-5/(-2)=2.5, so (2.5,20.25) is a candidate.
f4(x)=-x^2+7x+7, -b/(2a)=-7/(-2)=3.5, so (-3,-1) NO candidate.
f5(x)=2x^2+7x+5, -b/(2a)=-7/4=-1.75, so (-1.75,-1.125) is a candidate, and finally
f6(x)=-2x^2+8x+5, -b/(2a)=-8/(-4)=-2, so (-3,-1) NO candidate.

In principle, we have four candidates for four distinct parabolas, that should be the answer. However, it is better to check the value of f(x) to be sure.
For example,
f1(-3)=-1
f2(-4)=-4
f3(2.5)=20.25
f5(-1)=-1.125
So all candidates check out for the respective functions.

All that is left for you to do is to put the functions in the boxes corresponding to the value of x of the vertex.
For example, place f1(x) in the box corresponding to vertex (-4,-4), etc.)

Sorry, please post the piggy-backed question separately .

jcherry99 jcherry99 · Answer 2 · 2018-04-07T23:04:03+02:00

Let's start with an alternative method. One of the things you could do is graph the given equations. I've done the first 2 equations for you
Red Parabola
The red one is y = x^2 + 6x + 8
It is showing the lowest point as y = (x +3)^2 - 1 Just read this by completing the square.
You can get this answer by either completing the square or calculus
Blue Parabola
The blue parabola has it's minimum as shown on the graph as (-4,-4)
So it looks like this
y = a(x + 4)^2 - 4 That's the bare outline of the solution.
This one you have no choice but to complete the square, unless of course you just want to point in to solve for a
y = 2(x^2 + 8x + ) + 28
y = 2(x^2 + 8x + (8/2)^2 ) + 28 - 2*16
y = 2(x + 4)^2 - 4

Graphing it has the advantage of giving you the answer right away. If you have to show work, then you have to complete the square, but at least you know what the answer should be. If you have a graphing calculator so much the better. The program I did this with is desmos.