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Mathematics 14 Online
OpenStudy (anonymous):

Which of the following represents the graph of f(x) = 3x2 - 12 ?

OpenStudy (anonymous):

@Michele_Laino

OpenStudy (michele_laino):

from your equation, I get: \(f(0)=-12\), so the correct graph has to pass at point \((0,-12)\)

OpenStudy (michele_laino):

what are your options, please?

OpenStudy (anonymous):

There are 3 that do..

OpenStudy (michele_laino):

more precisely, the equation \(f(x)=3x^2-12\), is a parabola, which is concave up

OpenStudy (anonymous):

The first one lands on -2 and 2

OpenStudy (anonymous):

The next one lands on -3 and 3

OpenStudy (anonymous):

And the last one lands on -4 and 4

OpenStudy (michele_laino):

If I replace \(x=2\) I get: \(f(2)=3 \cdot 2^2-12=...?\)

OpenStudy (michele_laino):

please complete that computation

OpenStudy (anonymous):

0?

OpenStudy (anonymous):

3x4 is 12 - 12 is 0.

OpenStudy (michele_laino):

correct! Similarly, if I rplace \(x=-2\), I get: \(f(-2)=3 \cdot (-2)^2-12=...?\)

OpenStudy (michele_laino):

replace*

OpenStudy (anonymous):

Wouldn't that be 0 too?

OpenStudy (michele_laino):

correct!

OpenStudy (anonymous):

So how do I figure out which graph is right?

OpenStudy (michele_laino):

so, summarizing, we have these 3 conditions: 1) the parabola passes at point \((0,-12)\) 2) the parabola passes at point \((2,0)\) 3) the parabola passes at point \((-2,0)\) |dw:1447696487976:dw|

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