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anonymous

  • one year ago

Find a quadratic function f(x)=ax^2+bx+c for the parabola with minimum value -5 and x-intercepts -3 and 7??

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  1. anonymous
    • one year ago
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    You have to input the minimum value and the x,y coordinates into the vertex form of a parabola. Then convert the vertex form to standard form .

  2. anonymous
    • one year ago
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    ok thank you so much! @sophiav95

  3. campbell_st
    • one year ago
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    a really easy solution is to recognize the line of symmetry is halfway between the intercepts... and the vertex lies on the line of symmetry (-3 + 7)/2 = 2 so the line of symmetry is x = -2 you know the min value y = -5 so the vertex is (-2, -5) so then substitute the point into the vertex from of the parabola \[y = (x - h)^2 + k\] where (h, k) is the vertex

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