What could be the parabola with the following properties: minimum value of 2, no zeros and a y-intercept of 8.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The parabola that has to be determined has a minimum value of 2. It does not have any zeros and a y-intercept of 8.

As the parabola does not have any zeros it lies above the x-axis and opens upwards. There is no point of intersection of the parabola and the x-axis. Let the parabola be of the form y = ax^2 + bx + c

At the y-intercept, x = 0. This gives a*0^2 + b*0 + c = c = 8.

The minimum value of the function y = ax^2 + bx + 8 is 2. This is at the point where y' = 0

y' = 2ax + b = 0

=> x = -b/2a

At x = -b/2a, ax^2 + bx + 8 = 2

=> a(-b/2a)^2 + b(-b/2a) + 6 = 0

If we assume a = 1

=> b^2/4 -b^2/2 + 6 = 0

=> b^2/4 = 6

=> b^2 = 24

=> b = `sqrt 24` and b = `-sqrt 24`

A possible equation of the required parabola is `y = x^2 - sqrt 24 x + 8` and `y = x^2 + sqrt 24 x + 8 `

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