document.write( "Question 25761: Can you please help me with this, we are doing differential Calculus.
\n" ); document.write( "Find the equation of the tangetn for each given x-value.
\n" ); document.write( "y=x^2+8x+12, where x = -4.
\n" ); document.write( "At the back of the book, it says the answer is y=-4
\n" ); document.write( "I have no idea how to get that.
\n" ); document.write( "thank you.
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Algebra.Com's Answer #13875 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
As you probably know, the slope of the tangent at the given point (x = -4) is given by the value of the 1st derivative of the function \"y+=+x%5E2%2B8x%2B12\" at that point (x = -4).
\n" ); document.write( "\"dy%2Fdx+=+2x+%2B+8\" Substitute x = -4 and solve.
\n" ); document.write( "\"dy%2Fdx+=+2%28-4%29+%2B+8\"
\n" ); document.write( "\"dy%2Fdx+=+0\" So, the slope of the tangent is 0 which means that it's a horizontal line. To find the point (x, y) of tangency, you'll substitute x = -4 into the original equation.\r
\n" ); document.write( "\n" ); document.write( "\"y+=+%28-4%29%5E2+%2B+8%28-4%29+%2B+12\"
\n" ); document.write( "\"y+=+16+-+32+%2B+12\"
\n" ); document.write( "\"y+=+-4\"\r
\n" ); document.write( "\n" ); document.write( "The point of tangency is (-4, -4) and the equation of the line passing through that point with a slope of m = 0 is:
\n" ); document.write( " Using the point-slope form:\"y-y1+=+m%28x-x1%29\"
\n" ); document.write( "\"y+-%28-4%29+=+0%28x-%28-4%29%29\" Simplify.
\n" ); document.write( "\"y+%2B+4+=+0\"
\n" ); document.write( "\"y+=+-4\" This is the equation of the tangent at x = -4\r
\n" ); document.write( "\n" ); document.write( "Here's what it looks like in a graph:
\n" ); document.write( "\"graph%28300%2C200%2C-10%2C2%2C-6%2C5%2Cx%5E2%2B8x%2B12%2C-4%29\"
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