Identify the vertex for the graph of y = -2x2 - 12x + 1.

Question

yanasidlinskiy · Answer

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mathmale · Answer

First things first.  Your y = -2x^2 - 12x + 1 is a quadratic equation and has three constant coefficients:  a, b and c.  What are the values of a, b and c here?

Next, what is the formula for the x-coordinate of the vertex of a parabola?

anonymous · Answer

first coordinate of the vertex of 
$$y=ax^2+bx+c$$ is $-\frac{b}{2a}$

mathmale · Answer

rg:  would you please answer my question(s):  what are the values of a, b and c here?

anonymous · Answer

the values of a b and c are 
a=-2x^2
b=-12x and c=1

mathmale · Answer

Actually, rg, the coefficients are a=-2, b=-12 and c=1.  I was asking you to find those coefficients, without the power-of-x terms.


Now please take those values of a and b and substitute them into the formula for the x-coordinate of the vertex.

mathmale · Answer

See satellite73's post, above, for that formula.

anonymous · Answer

okay so it would be -12/2(-2) which is -12/-4 and that equals 3 right?

mathmale · Answer

$$x _{vertex}=\frac{ -b }{ 2a }\rightarrow x _{v}=\frac{ -(-12 ) }{ 2(-2) }$$=24/(-4) = ?

anonymous · Answer

thats -6

mathmale · Answer

But wouldn't -(-12) be +12?
Wouldn't 2(-2) be -4?

anonymous · Answer

uhm yeah but then that would make it -3

mathmale · Answer

Yes, that's right.  Now, if you want to find the vertex, take that $$x _{vertex}=-3$$and substitute it for x in your function, $$f(x)=-2x^2-12x+1$$to find the y-coordinate of the vertex.  
 Last, write out the point that represents the vertex:  It will look like (-3,    ??     ).

mathmale · Answer

Sorry, but I'm getting off OpenStudy now.  I believe you have enuf info with which to finish this problem.   If not, you could always ask for help again.  Good luck!