document.write( "Question 1986: Sorry, long time since I've done maths! The question is \"find a formula for the quadratic function as shown.\" The intercepts are (x= -1,y=2,) ( x=1, y=0 )and (x= -3, y=0.) Thanks for your time and help. \n" ); document.write( "
Algebra.Com's Answer #798 by longjonsilver(2297)\"\" \"About 
You can put this solution on YOUR website!
rightio..i understand now :-)\r
\n" ); document.write( "\n" ); document.write( "however, the point (-1, 2) is not an intercept, it is the maximum (or minimum) point on the curve...\r
\n" ); document.write( "\n" ); document.write( "ok. where an equation/line/curve crosses the x-axis is where y=0. Finding such values (called the roots of the equation) is the main aim in algebra like this.\r
\n" ); document.write( "\n" ); document.write( "typically we have a quadratic equation equal to zero which we factorise because then we have 2 parts that multiply to equal zero. For this to be true, one of those 2 parts must be zero itself. Which one? either.\r
\n" ); document.write( "\n" ); document.write( "so here, we work backwards...we know that the roots are x=1 and x=-3, so we know that\r
\n" ); document.write( "\n" ); document.write( "(x-1)(x+3) = 0
\n" ); document.write( "so, now we can multiply this out to give \"x%5E2+%2B+3x+-x+-+3+=+0\". Collecting the 2 middle x-terms gives \"x%5E2+%2B+2x+-+3+=+0\" -eqn1\r
\n" ); document.write( "\n" ); document.write( "so what is the equation? well it is \"y+=+x%5E2+%2B+2x+-+3\", but hold on...what about moving all the terms in eqn1 to the other side so we had \"0+=+-x%5E2+-+2x+%2B+3\"? Here we have the equation \"y+=+-x%5E2+-+2x+%2B+3\"\r
\n" ); document.write( "\n" ); document.write( "So we have 2 quadratics that both go through the 2 roots mentioned. Here they are sketched...\r
\n" ); document.write( "\n" ); document.write( "\"graph%28300%2C200%2C-8%2C4%2C-8%2C8%2Cx%5E2+%2B+3x+-x+-+3%2C+-x%5E2+-+2x+%2B+3%29\"\r
\n" ); document.write( "\n" ); document.write( "As you can see, they are both symmetric about the max/min axis (at x=-1). So we need this third point to tell us which of the 2 curves, and hence, which of the 2 equations we had.\r
\n" ); document.write( "\n" ); document.write( "Note:
\n" ); document.write( "a u-shape quadratic has +ve x-squared term
\n" ); document.write( "a n-shape quadratic has -ve x-squared term\r
\n" ); document.write( "\n" ); document.write( "so, the point (-1, 2) should lie on the n-shaped curve and hence our equation was \"y+=+-x%5E2+-+2x+%2B+3\". However, looking at the curve, the point is more like (-1, 4). Oh well!\r
\n" ); document.write( "\n" ); document.write( "Hope this was of help to you.\r
\n" ); document.write( "\n" ); document.write( "cheers
\n" ); document.write( "Jon.
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