SOLUTION: A quadratic equation has an x intercept at 2.5 and 5 and a y value y=-2.25. Use any valid method to calulate the equation of the quadratic in the form y=ax^2 +bx+c
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-> SOLUTION: A quadratic equation has an x intercept at 2.5 and 5 and a y value y=-2.25. Use any valid method to calulate the equation of the quadratic in the form y=ax^2 +bx+c
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Question 893825: A quadratic equation has an x intercept at 2.5 and 5 and a y value y=-2.25. Use any valid method to calulate the equation of the quadratic in the form y=ax^2 +bx+c Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! A quadratic equation has an x intercept at 2.5 and 5 and a y value y=-2.25. Use any valid method to calulate the equation of the quadratic in the form y=ax^2 +bx+c
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(x-2.5)(x-5)=-2.25
x^2-7.25x+12.5=-2.25
x^2-7.25x+14.75=0
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