Question 71903
a. First complete the square
{{{y=5x^2-10x+3}}}
{{{y-3=5x^2-10x}}}
{{{y-3+5=5(x^2-2x+1)}}}Add 5(-2/2)^2 to both sides to complete the square
{{{y+2=5(x-1)^2}}}Factor the right side
{{{y=5(x-1)^2-2}}}There's the completed square.
In the basic equation
{{{y=a(x-h)^2+k}}}} h is the x-coordinate of the vertex and k is the y-coordinate of the vertex. So in
{{{y=5(x-1)^2-2}}} the vertex is (1,-2) you can verify this by graphing
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b. The axis of symmetry is at the vertex, so the axis of symmetry is x=1
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c. The min or max of any quadratic is always at the vertex. So in this case the min is at (1,-2) (the min is the lowest point on the curve)
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d. The range is all of the outputs of the function (or all the y-values). So the range is all the y's from -2 to infinity or could be written as [-2,infinity)
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e. Both {{{y=5(x-1)^2-2}}} and {{{y=5x^2-10x+3}}} have the same graphs (because they are equivalent)
{{{graph( 300, 200, -5, 5, -5, 5, 5x^2-10x+3)}}}