How do you write the equation of the quadratic function with roots 6 and 10 and a vertex at (8, 2)?

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Answer 1 · 2015-06-16T23:12:03

In vertex form:

#y = -1/2(x-8)^2+2#

In standard form:

#y = -1/2x^2+8x-30#

Since we're given the vertex the equation may be written

#y = a(x-8)^2 + 2# for some constant #a#.

Putting #x=10#, #y=0# into this equation, we find:

#0 = a(10-8)^2 + 2 = 4a+2#

Hence #a = -1/2#

So in vertex form, the equation is:

#y = -1/2(x-8)^2 + 2#

Expand to get standard form:

#y = -1/2x^2+8x-30#