document.write( "Question 176168: this is one question with different parts
\n" ); document.write( "Find the vertex, line of symmetry, maximum or minimum value of the quadratic function, and graph the function.
\n" ); document.write( "f(x)=-2x^2+2x+3
\n" ); document.write( "x-coordinate of vertex
\n" ); document.write( "y-coordinate of vertex
\n" ); document.write( "equation of line of symmetry
\n" ); document.write( "max/min value of f(x)
\n" ); document.write( "The value of f(1/2)=7/2 is min or max
\n" ); document.write( "Please help I have no idea what I am doing thank you
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Algebra.Com's Answer #131271 by MathLover1(20849)\"\" \"About 
You can put this solution on YOUR website!
1. Quadratic into Vertex Form\r
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Solved by pluggable solver: Completing the Square to Get a Quadratic into Vertex Form

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\n" ); document.write( " \"y=-2+x%5E2%2B2+x%2B3\" Start with the given equation
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\n" ); document.write( " \"y-3=-2+x%5E2%2B2+x\" Subtract \"3\" from both sides
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\n" ); document.write( " \"y-3=-2%28x%5E2-1x%29\" Factor out the leading coefficient \"-2\"
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\n" ); document.write( " Take half of the x coefficient \"-1\" to get \"-1%2F2\" (ie \"%281%2F2%29%28-1%29=-1%2F2\").
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\n" ); document.write( " Now square \"-1%2F2\" to get \"1%2F4\" (ie \"%28-1%2F2%29%5E2=%28-1%2F2%29%28-1%2F2%29=1%2F4\")
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\n" ); document.write( " \"y-3=-2%28x%5E2-1x%2B1%2F4-1%2F4%29\" Now add and subtract this value inside the parenthesis. Doing both the addition and subtraction of \"1%2F4\" does not change the equation
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\n" ); document.write( " \"y-3=-2%28%28x-1%2F2%29%5E2-1%2F4%29\" Now factor \"x%5E2-1x%2B1%2F4\" to get \"%28x-1%2F2%29%5E2\"
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\n" ); document.write( " \"y-3=-2%28x-1%2F2%29%5E2%2B2%281%2F4%29\" Distribute
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\n" ); document.write( " \"y-3=-2%28x-1%2F2%29%5E2%2B1%2F2\" Multiply
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\n" ); document.write( " \"y=-2%28x-1%2F2%29%5E2%2B1%2F2%2B3\" Now add \"3\" to both sides to isolate y
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\n" ); document.write( " \"y=-2%28x-1%2F2%29%5E2%2B7%2F2\" Combine like terms
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\n" ); document.write( " Now the quadratic is in vertex form \"y=a%28x-h%29%5E2%2Bk\" where \"a=-2\", \"h=1%2F2\", and \"k=7%2F2\". Remember (h,k) is the vertex and \"a\" is the stretch/compression factor.
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\n" ); document.write( " Check:
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\n" ); document.write( " Notice if we graph the original equation \"y=-2x%5E2%2B2x%2B3\" we get:
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\n" ); document.write( " \"graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C-2x%5E2%2B2x%2B3%29\" Graph of \"y=-2x%5E2%2B2x%2B3\". Notice how the vertex is (\"1%2F2\",\"7%2F2\").
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\n" ); document.write( " Notice if we graph the final equation \"y=-2%28x-1%2F2%29%5E2%2B7%2F2\" we get:
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\n" ); document.write( " \"graph%28500%2C500%2C-10%2C10%2C-10%2C10%2C-2%28x-1%2F2%29%5E2%2B7%2F2%29\" Graph of \"y=-2%28x-1%2F2%29%5E2%2B7%2F2\". Notice how the vertex is also (\"1%2F2\",\"7%2F2\").
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\n" ); document.write( " So if these two equations were graphed on the same coordinate plane, one would overlap another perfectly. So this visually verifies our answer.
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Solved by pluggable solver: Min/Max of a Quadratic Function
The min/max of a quadratic equation is always at a point where its first differential is zero. This means that in our case, the value of \"x\" at which the given equation has a maxima/minima must satisfy the following equation:\"2%2A-2%2Ax%2B2=0\"
\n" ); document.write( " => \"x+=+-2%2F%282%2A-2%29+=+0.5\"
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\n" ); document.write( " This point is a minima if value of coefficient of x2 is positive and vice versa. For our function the point x=0.5 is a \"maxima\" The graph of the equation is :
\n" ); document.write( "\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+-2%2Ax%5E2%2B2%2Ax%2B3+%29\"
\n" ); document.write( " Alternate method
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\n" ); document.write( " In this method, we will use the perfect square method.
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\n" ); document.write( " Step one:
\n" ); document.write( " Make the coefficient of \"x%5E2\" positive by multiplying it by \"-1\" in case\"a%3C0\".
\n" ); document.write( " Maxima / Minima is decided from the sign of 'a'.
\n" ); document.write( " If 'a' is positive then we have Minima and for 'a'negative we have Maxima.
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\n" ); document.write( " Step two:
\n" ); document.write( " Now make the perfect square with the same \"x%5E2\" and \"x\" coefficient.
\n" ); document.write( " \"%281.4142135623731%2Ax+%2B+%28-2%2F2%29%29%5E2\"
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\n" ); document.write( " Maxima / Minima lies at the point where this squared term is equal to zero.
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\n" ); document.write( " Hence,
\n" ); document.write( " =>\"x=%28-%28-2%2F2%29%2F1.4142135623731%29=+0.707106781186547\"
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\n" ); document.write( " This point is a minima if value of coefficient of x2 is positive and vice versa. For our function the point x=0.5 is a \"maxima\".
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\n" ); document.write( " For more on this topic, refer to Min/Max of a Quadratic equation.

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