Question 159802
you can find it in a couple of ways.
the easiest is to remember that the vertex is given by the equation {{{x = -b/(2*a)}}}.
how do you find this?
your original equation is a quadratic equation.
the general form of a quadratic equation is {{{y = a*x^2 + b*x + c}}}
your equation is {{{y = x^2 + 3*x - 4}}}
this is the same as {{{y = 1*x^2 + 3*x - 4}}}
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
{{{x = -3/(2*a)}}}
which becomes
{{{x = -3/(2*1)}}}
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 {{{y = x^2 + 3*x - 4}}} becomes {{{y = (-3/2)^2 + 3*(-3/2) - 4}}} becomes {{{y = 2.25 + (-4.5) - 4}}} becomes {{{y = -2.25 - 4}}}
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.
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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.
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outside of the equation for the vertex ({{{x = -3/(2*a)}}}), 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).
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going back to the general form of the equation ({{{y = a*x^2 + b*x + c}}}),
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.
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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.
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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.
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the graph of the equation is {{{y = x^2 + 3*x - 4}}}
the graph looks like this.
please scan below the graph for further comments.
{{{graph(600,600,-15,15,-10,100,x^2+3*x-4)}}}
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 {{{x = -b/(2*a)}}} 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 {{{y = a*x^2 + b*x + c}}}
formula for x-value of vertex is {{{x = -b/(2*a)}}}
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.