If I have 2 points (let’s say $(0,8)$ and $(6,0)$) and a line on which a vertex can be ($f(x)=16$), how can I find the possible quadratic equations that would intersect both the points and have the vertex lay on that line?
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A quadratic $y = ax^2 + bx + c$ has three parameters that define it: $a,b,c$. The three facts you know about the quadratic each provide an equation in those parameters. Three equations, three unknowns. You can solve it. – Paul Sinclair Jan 24 '23 at 12:58
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Let the vertex be V(h, 16). Then the quadratic curve is $y = a(x - h)^2 + 16$ for some negative value of $a$ and also $h$.
All this deduction comes from the knowledge of the locations P(0, 8), Q(6, 0), and V.
Since it must pass through P, then $8 = a(0 - h)^2 + 16$.
Similarly for Q, we have $0 = a(6 - h)^2 + 16$
Solving the two, we get $h^2 + 12h -36 = 0$, from which we get h = -14.48528137 after rejecting the other impossible value.
Mick
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