Question

Which graph shows the quadratic function y = 3x2 + 12x + 14?

Answer 1

@jdoe0001 can you please help me. I don't understand what I need to do please. :s

Answer 2

I know the answer isn't going to be D or C but i can't figure it out from there pleaseeeeeeee help. :s

Answer 3

so it is C?

Answer 4

yeah.

Answer 5

You could always go with the old school approach of actually computing some values and looking at the graph! x=0 is a good candidate because it is trivial to compute. \[y = 3x^2+12x+14 = 3(0)^2+12(0)+14 = 14\]Any graph not containing a curve going through \((0,14)\) can immediately be rejected. Unfortunately, none of the graphs actually show the y-intercept, probably to prevent this approach from being so easy :-)

Next choice is x = 2 which certainly does appear on two of the graphs:
\[y = 3(2)^2 + 12(2) + 14 = 12+24+14 = 50\]We can reject B and D because they do have a point at \(x =2\) and it isn't \(y=50\)

To decide between A and C, you could do x = -1 or x = -2, -1 is a bit easier:\[y = 3(-1)^2+12(-1) + 14 = 3-12+14 = 5\]Graph C is the one that goes through the point (-1,5) so it is our choice.

Answer 6

Another approach, if you know anything about parabolas, is to find the vertex, and see which graph contains a parabola with a vertex there.

In an equation of the form \(y = ax^2+bx+c\) the vertex has an x-coordinate of \[x=-\frac{b}{2a}\]Our quadratic is \(y = 3x^2+12x+14\) so we have \(a=3,b=12,c=14\)
\[x = -\frac{12}{2*3} = -2\]Pick the graph with a parabola having a vertex at x = -2. Again, C is our choice.

Answer 7

thanks a bunch :)

Answer 8

the answers right btw :)