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Let's try another one:

Let's graph y = -2x^2 - 4x + 1, label 3 points (like we 
did before) and find where it crosses the x-axis (these are called "zeros"):

 

1Complete the square:

y = -2x^2 - 4x + 1 gives y = -2 ( x^2 - 4x ___ ) + 1 ___ which gives y = -2 ( x^2 - 4x + 1 ) + 1 + 2 which gives y = -2 ( x + 1 )^2 + 3

2Graph it and label three points:

y = -2 ( x + 1 )^2 + 3 ... upside down, 2 times as tall, shifted 1 to the left and up 3

graph of y = -2 ( x + 1 )^2 + 3

3Find the zeros:

*set it =0 and solve

-2x^2 - 4x + 1 = 0

We'll have to quadranate this one:

x = 4 +/- sqrt( ( -4 )^2 - 4 ( -2 ) ( 1 ) ) / ( 2 ( -2 ) ) = 4 +/- sqrt( 16 + 8 ) / -4 which gives x = 4 - sqrt( 24 ) / -4 or x = 4 + sqrt( 24 ) / -4 ... ( x is approximately equal to .22 or -2.22 )

Do these values make sense with our graph?

graph of -2x^2 - 4x + 1 = 0 ... the x-intercepts are x = 4 - sqrt( 24 ) / -4 and x = 4 + sqrt( 24 ) / -4

YES!