Question 438654: The question asks students to graph y=-x^2+4x+1 and to have all x intercepts and y intercepts labeled. I have figued out that the vertex is (2,5), the y intercept is (0,1) and I have worked out the problem to get points of (2,5) & (-1,-4) I am having an issue figurng out exactly what the points are when the lines cross the x axis. Thank you for your help.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! The question asks students to graph y=-x^2+4x+1 and to have all x intercepts and y intercepts labeled. I have figued out that the vertex is (2,5), the y intercept is (0,1) and I have worked out the problem to get points of (2,5) & (-1,-4) I am having an issue figurng out exactly what the points are when the lines cross the x axis.
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y=0 when the parabola crosses the x-axis.
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Solve -x^2+4x+1 = 0
Use the Quadratic Formula:
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x = [-4 +- sqrt(16-4*-1*1)]/(-2)
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x = [-4 +- sqrt(20)]/(-2)
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x = [-4 +- 2sqrt(5)]/(-2)
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x = [2+- sqrt(5)]
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The curve crosses the x-axis at 2-sqrt(5) and at 2+sqrt(5)
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Cheers,
Stan H.
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