Question 245946: Quadratic Functions and Equations
The first question I gave it go at.
Solve.
3t^2-2=0
+2+2
3t^2=2
______
3 3
t^2=2/3 or (I don't know how to do the square root sign on here.)
_ ______ _ _ __________
V2/3 or V 2/3
Hope I did it right.
Then the next is also solve.
x^2+7=3x
Now so far all I have done is taken 3x from both sides
which brings us to
x^2-3x+7=0
At this point I am lost. Because normally I would think we would do this but I can't find anything to add up to equal 3
(x )(x )=0
So I am stuck with this problem.
Last problem I am having is one that I got so confused w/ the process that I couldn't even start it. I am just lost.
It states:
For each quadratic function, (a) find the vertex and the axis of symmetry and (b) graph the function.
g(x)=2x^2+5x-1
I know we follow the form of a(x-h)^2+k. But I am so lost at this. I know it is most likely real easy but I am lost.
Found 2 solutions by richwmiller, solver91311:
Answer by richwmiller(17219)   (Show Source):
You can put this solution on YOUR website! you should only have one problem per entry.
Sqrt is the symbol for square root
So you should have plus or minus Sqrt (2/3) is correct
x^2+7=3x
Now so far all I have done is taken 3x from both sides
which brings us to
x^2-3x+7=0
At this point I am lost. Because normally I would think we would do this but I can't find anything to add up to equal 3
so you have to use another method
the quadratic formula or complete the square
I will show you complete the square.
so we rearrange the equation.
x^2-3x=-7
take half of -3 and square it
(3/2)^2=9/4
add 9/4 to both sides
x^2-3x+9/4=-28/4+(9/4)=-19/4
X^2-3x+9/4=-19/4
factor the left side
(x-3/2)^2=-19/4
now we see we will be dealing with imaginary numbers since we need to get the sqrt of a minus number
get he sqrt of both sides
|x-3/2|=(i*sqrt(19))/2
x=3/2+-(i*sqrt(19))/2
x=1/2(3-i*sqrt19)
x=1/2(3+i*sqrt19)
Answer by solver91311(24713)  (Show Source):
You can put this solution on YOUR website!
Hint for next time: show square roots like this: sqrt(x) whatever is inside the square root sign is inside the parentheses. show plus or minus like this: +\-
So you got yourself to:
and then you took the square root of both sides to get:
So far, so good. But, you need to simplify further. A radical in a denominator is not simplest form. First apply the rule that the square root of a quotient is the quotient of the square roots.
So:
Now multiply your fraction by 1 in the form of 
\ =\ \pm\frac{\sqrt{6}}{3})
And now you are in simplest form, though you might not think so just to look at it. But the rule is: No matter how ugly the numerator gets, a fraction is simpler if it has no radical in the denominator.
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You are right. There are no integer factors for the given quadratic. So you need to either Complete the Square (which will get messy because the first degree coefficient is odd), or use the quadratic formula:
where  ,  , and  are the high order, first degree, and constant coefficients of the quadratic in standard form. For your problem:  ,  , and 
Since the discriminant,  , the solution will be a conjugate pair of complex roots of the form  where  is the imaginary number defined as 
\ \pm\ \sqrt{(-3)^2\ -\ 4(1)(7)}}{2(1)})
===========================================
For any function of the form
\ =\ ax^2\ +\ bx\ +\ c)
the  -coordinate of the vertex is given by 
The  -coordinate of the vertex is the value of the function at the -coordinate of the vertex, namely: )
This is a parabola, so there is only one line of symmetry, namely the vertical line that passes through the vertex, so the equation is:
If the lead coefficient is positive, then the parabola opens upward. If the parabola opens upward then quite obviously the -coordinate of the vertex is the minimum of the function. In this case, the function has no maximum.
If the lead coefficient is negative, then the opposite is true: The parabola opens downward, the value of the function at the vertex is a maximum, and there is no minimum.
So, for your particular problem, you need to first calculate the value of }{2(2)}) , and then calculate the value of the function at that value:
}{2(2)}\right)\ =\ 2\left(\frac{-(5)}{2(2)}\right)^2\ +\ 5\left(\frac{-(5)}{2(2)}\right)\ -\ 1)
That will give you the vertex and the axis of symmetry.
Evaluate
\ =\ 2\left(0\right)^2\ +\ 5\left(0\right)\ -\ 1)
To determine the point \right)) where the graph intersects the  -axis.
Solve the equation:
to find the points where the graph intersects the  -axis.
Use symmetry to find other points. If  is the -coordinate of the vertex, \ =\ f\left(x_v\ -\ a\right))
Plot all of your points and draw a smooth curve.
John
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