Question 110446
Let's start with what the graph SHOULD look like and go from there.
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{{{graph(400,400,-7,7,-7,7,x^2+3x-4)}}}
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Next, let's find the roots of the equation {{{y=x^2+3x-4}}}
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Looking at the value of the b and c coefficients, the first thing we can tell is that our two factors are going to be of the form {{{(x-a)(x+b)}}}.  That's because the sign on the constant term is negative.  The sign on the first degree term (3x) being positive tells us that the absolute value of b has to be larger than the absolute value of a in {{{(x-a)(x+b)}}}.  So what are the possible values for a and b?
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+/- 1, +/-2, and +/-4
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Looking at that, -1 and +4 have a sum of +3, so that looks like our numbers.
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{{{0=x^2+3x-4}}}
{{{0=(x-1)(x+4)}}}
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Which means that {{{x-1=0}}} or {{{x+4=0}}}, which is to say {{{x=1}}} or {{{x=-4}}}.  Looking at the intercepts you posted, you can see that your signs are reversed.  This may be one source of your difficulties.
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Looking at the graph, your y-intercept seems to be correct, but let's check by determining the value of {{{y=x^2+3x-4}}} when {{{x=0}}}.  By inspection you should see that {{{y=-4}}}, so that intercept point you had was correct.
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You didn't share the value of the coordinates you found for the vertex, but the vertex, (h,k), should have an x-coordinate of {{{h=-b/2a}}}, in this case {{{-3/2}}}. Substitute in the orginial equation to get the y-coordinate of the vertex.  In this case, 
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{{{k=h^2+3h-4}}}
{{{k=(-3/2)^2+3(-3/2)-4}}}
{{{k=-6.25}}}.
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Having determined these points, you should then be able to select some candidate x values to determine a few more points.  I suggest you use 2, -1, -2, -3, and -5.  That should give you enough points, assuming careful arithmetic, to plot so that you can draw a smooth curve.