document.write( "Question 110446: I am trying to graph a quadratic equation....everything was going fine until I tried to get the points for the left side of the graph....they're all over the place...some go below the vertex and it definatly doesnt look like a parabola. Can someone please explain to me how to get the -points? I tried plugging x into the equation but its not working at all...neither are the x and y intercepts, which I have as y=-4 and x=4 and -1, I'm getting so frustrated. The equation is X²+3x-4
\n" ); document.write( " I appreciate anyones help!
\n" ); document.write( " Bonnie C. \n" ); document.write( "
Algebra.Com's Answer #80494 by solver91311(24713)\"\" \"About 
You can put this solution on YOUR website!
Let's start with what the graph SHOULD look like and go from there.
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\n" ); document.write( "\"graph%28400%2C400%2C-7%2C7%2C-7%2C7%2Cx%5E2%2B3x-4%29\"
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\n" ); document.write( "Next, let's find the roots of the equation \"y=x%5E2%2B3x-4\"
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\n" ); document.write( "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 \"%28x-a%29%28x%2Bb%29\". 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 \"%28x-a%29%28x%2Bb%29\". So what are the possible values for a and b?
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\n" ); document.write( "+/- 1, +/-2, and +/-4
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\n" ); document.write( "Looking at that, -1 and +4 have a sum of +3, so that looks like our numbers.
\n" ); document.write( ":
\n" ); document.write( "\"0=x%5E2%2B3x-4\"
\n" ); document.write( "\"0=%28x-1%29%28x%2B4%29\"
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\n" ); document.write( "Which means that \"x-1=0\" or \"x%2B4=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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\n" ); document.write( "Looking at the graph, your y-intercept seems to be correct, but let's check by determining the value of \"y=x%5E2%2B3x-4\" when \"x=0\". By inspection you should see that \"y=-4\", so that intercept point you had was correct.
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\n" ); document.write( "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%2F2a\", in this case \"-3%2F2\". Substitute in the orginial equation to get the y-coordinate of the vertex. In this case,
\n" ); document.write( ":
\n" ); document.write( "\"k=h%5E2%2B3h-4\"
\n" ); document.write( "\"k=%28-3%2F2%29%5E2%2B3%28-3%2F2%29-4\"
\n" ); document.write( "\"k=-6.25\".
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\n" ); document.write( "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.
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