SOLUTION: Hi, I'm having trouble with the following question... Find the x- and y- coordinates of the parabola vertex for the function f(x)=-3x^2+6x-1 My textbook tells me to + and

Question 273706: Hi,
I'm having trouble with the following question...
Find the x- and y- coordinates of the parabola vertex for the function
f(x)=-3x^2+6x-1
My textbook tells me to + and - the coefficient of x
So I do this...
f(x)=-3x^2+6x-1
f(x)=-3x^2+6x+3-3-1
f(x)=(-3x^2+6x+3)-3-1
And now I need to factor and simplify
any help appreciated.
thank you,
Found 2 solutions by solver91311, Theo:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!

I'm not sure what you are trying to say as to your textbook instructions.

However, here is how to find the vertex of a parabola when you have a quadratic function in standard form:

The -coordinate of the vertex is given by:

and then the -coordinate of the vertex is given by:

So for your problem:

And

You can do your own arithmetic.

And look at a graph of your function to check your work:

John

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
equation is:

f(x)=-3x^2+6x-1

The vertex is at x = -b/2a

since this equations is in the standard form of ax^2 + bx + c, then:

a = -3
b = 6
c = -1

vertex is at x = -b/2a = -6/-6 = 1.

the y coordinate is at y = f(1) = -3*1^2 + 6*1 - 1 = -3 + 6 -1 = 2

the vertex should be at (x,y) = (1,2)

graph of this equation is shown below:

looking at the graph, this appears to be accurate.

the vertex is the max/min point and it is at (x,y) = (1,2) as predicted by the formula.

I'm not exactly sure of what your textbook is telling you and why.

The vertex is found by the formula I just showed you if this is a quadratic equation.

If you are looking for the roots of the equation, those roots can be found by factoring the equation or by the use of the quadratic formula.

the equation is, once again, -3x^2 + 6x - 1

I don't believe the roots can be found by factoring, so you would have to resort to the quadratic formula or the completing the squares method.

The roots of the equation are the points at which the graph crosses the x-axis.

You can see from the graph that those roots should be around .1 and 1.8.

I could solve for them, but I believe the questions was to find the vertex, which I did for you above.

The quadratic formula to find the roots is:

as before, the equation is in standard form of f(x) = -3x^2 - 6x + 1 and:

a = -3
b = -6
c = 1

You just plug those values into the quadratic formula and you find the x value of your roots, if the values are real.

If the values are imaginary or complex, then the roots are not real and the graph of the equation will not cross the x-axis.

This equation has real roots as can be seen from the graph.

I have no idea what your textbook is telling you.

If the equation is symmetric than f(x) = f(-x) which means that you have an axis of symmetry and your graph will have values to the left of that axis and to the right of that axis.

But that's another topic.

The answer to your question (I think) is that the vertex of the equation is at the point (x,y) = (1,2) as can be confirmed by looking at the graph.

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