What graph represents the function y = 3x^2 + 12x - 6?

Question

cloverracer · Answer

i know the parabola would be opening upward.

mathmale · Answer

Thank you for stating what you already know.  You're right; this parabola will open upward.  Were there illustrations with this problem?

whpalmer4 · Answer

Yes, the coefficient of $x^2$ is positive, so the parabola opens upward.

With a parabola in the form $y = ax^2+bx+c$, we can find the x-coordinate of the vertex of the parabola by evaluating $$x = -\frac{b}{2a}$$
That may give you enough information to solve your problem without showing us the graphs.

cloverracer · Answer

yes I used process of elimination, and got down to 2 illustrations. I will post them

whpalmer4 · Answer

The y-coordinate can be gotten by evaluating the function at that value of $x$

whpalmer4 · Answer

Alternatively, you could rearrange the formula into "vertex form", which is $$y = a(x-h)^2+k$$where the vertex of the parabola is at $(h,k)$
This probably isn't the direct route to take for this problem, but I provide it for completeness' sake if you're taking notes for future reference.

cloverracer · Answer

attachment

cloverracer · Answer

I used process of elimination and got down to these illustrations.

whpalmer4 · Answer

So, given my first suggestion, what is the x-coordinate of the vertex of the parabola described by $$y = 3x^2+12x-6$$That's all you need to make the correct choice here.

cloverracer · Answer

for the x-coordinate I got x = -2

cloverracer · Answer

ohh! nevermind I got it. So it would be the first one? because the x-coordinate is -2 and the line of symmetry of the parabola goes through -2 and the vertex.

cloverracer · Answer

@mathmale

whpalmer4 · Answer

Yes, that is correct.

cloverracer · Answer

okay thank you!!

whpalmer4 · Answer

Yet another approach to telling which of these two parabolas is the correct choice would be to find the solutions where y = 0.  Not quite so easy in this case as you can't factor it, but you could complete the square or use the quadratic formula.

whpalmer4 · Answer

You could eyeball the graph and say that the first one crosses the x-axis at about x = 1/2, and plug that in to see if you get a y-value that is in the ballpark:

$$3(\frac{1}{2})^2 + 12(\frac{1}{2}) -6 = 3*\frac{1}{4}+6-6 = \frac{3}{4}$$that's pretty close; the other graph has y = -8 or so at x = 1/2.  The actual x value where y = 0 is $x = \sqrt{6}-2 \approx 0.44949$ which is why it didn't come out to be exactly 0.