SOLUTION: I am trying to find the vertex for the following problem: x^2+3x-4=0 All the help I could get would be appreciated.

Question 159802: I am trying to find the vertex for the following problem:
x^2+3x-4=0
All the help I could get would be appreciated.

Found 2 solutions by scott8148, gonzo:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
the vertex is on the axis of symmetry

general eqn for axis is x=-b/(2a)

x=-3/(2*1) __ x=-3/2

substituting __ y = (-3/2)^2 + 3(-3/2) - 4 __ y = 9/4 - 9/2 -4 __ y = -25/4

vertex is (-3/2,-25/4)
Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
you can find it in a couple of ways.
the easiest is to remember that the vertex is given by the equation .
how do you find this?
your original equation is a quadratic equation.
the general form of a quadratic equation is
your equation is
this is the same as
if you compare your equation to the general form of the equation, you'll see that
a = 1
b = 3
c = -4
plugging these values in into the equation for the vertex, we get

which becomes

which becomes
x = -3/2
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that finds the x value for the vertex.
to find the y value for the vertex, plug the value of x for the vertex into the equation and solve.
the equation of becomes becomes becomes
which becomes y = -6.5
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the vertex for this equation becomes (-3/2, -6.5).
the x value is -3/2.
the y value is -6.5.
-----------------------------
since the vertex is the point in the graph where the y value is either a maximum or a minimum, we can assume that this graph will change direction at that point.
------------------------------
outside of the equation for the vertex (), the next best thing to know is whether the quadratic equation is pointing upwards (vertex is a maximum) or whether the quadratic equation is pointing downwards (vertex is a minimum).
-------------------------------
going back to the general form of the equation (),
the graph is pointing upwards (vertex is maximum value for y) when a is negative, and the graph is pointing downwards (vertex is minimum value for y) when a is positive.
---------------------------------
this should be what you would think would happen intuitively.
if y = x^2, then for negative values of x or positive values of x, y = x^2 will always be positive. that tells you that the graph will be pointing downwards and the vertex will be a minimum.
if y = -x^2, then for negative values of x or positive values of x, y = -x^2 will always be negative. that tells you that the graph will be pointing upwards and the vertex will be a maximum.
---------------------------------
in the equation provided, a = 1 which is positive.
we should expect then that the graph is pointing downwards and that the vertex of (-3/2,-6.5) is a minimum value.
----------------------------------
the graph of the equation is
the graph looks like this.
please scan below the graph for further comments.

as you can see, the graph is point downwards and the vertex is the minimum.
this is because a was positive (a*x^2 in the general form of the equation). specifically, a was +1.
you can also see that the minimum is where we calculated it to be.
x = -3/2 is the same as -1.5.
y = -6.25.
vertex = (-1.5,-6.25)
x value of the vertex was provided by the equation where a was 1, and b was 3.
y value of the vertex was provided by plugging x value of -3/2 into the equation and solving.
things to remember:
general form of quadratic equation is
formula for x-value of vertex is
y-value of vertex is found by plugging x-value of vertex into the equation to be solved.
when a is positive, equation points downward.
when a is negative, equation points upward.
when the equation points downward, the vertex is a minimum value of y.
when the equation points upward, the vertex is a maximum value of y.

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