Answer:
<em>x^2+5x-6=3</em>
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<em>1. the parabola opens up as the leading coefficient is a positive.</em>
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<em>2. to find the vertex to find the x value</em>
<em>-5/2(1)= -5/2 (-2 1/2) (-2.5)</em>
<em>to find the y value of the vertex substitie the x value into the equation</em>
<em>(-2 1/2)^2+5(-2 1/2)-6=3</em>
<em>6 1/4 - 12 1/2 -9 = y</em>
<em>-15 1/4 = y</em>
<em>y = -15 1/4 (-15.25)</em>
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<em>(-2.5, 15.25) , (-2 1/2, -15 1/4)</em>
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<em>3. the y intercept can be found by substituting the X's with 0's</em>
<em>x^2+5x-6=3</em>
<em>first move the terms to the left</em>
<em>x^2+5x-9=y</em>
<em>(0)^2+5(0)-9=y</em>
<em>0+9=y</em>
<em>9=y</em>
<em>y=9</em>
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<em>the y intercept is 9 (0,9)</em>
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<em>4. b^2-4ac</em>
<em>5^2-4(1)(-9)</em>
<em>=25+36</em>
<em>=61</em>
<em>61 is greater than 0 so this means there are 2 real solutions.</em>
<em>the +- next to each other is meant to be plus or minus </em>
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<em>6. to find three other points pick any x value and substitute that into the equation</em>
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<em>y=(2)^2+5(2)-9</em>
<em>y=4+10-9</em>
<em>y=5</em>
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<em>(2,5)</em>
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<em>y=(3)^2+5(3)-9</em>
<em>y=9+15-9</em>
<em>y=15</em>
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<em>(3,15)</em>
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<em>y=(4)^2+5(4)-9</em>
<em>y=16+20-9</em>
<em>y=27</em>
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<em>(4,27)</em>