Find the quadratic equation for a parabola that passes through $$(1,0) (5,0) (0,10)$$
To do this I turned it into $$ x = 1 $$ $$ x = 5 $$ and then into $$(x-1)(x-5)$$ after you multiply everything it turns into $$Y = a(x^2-6x+5)$$ Now My question is how would I find A? Do i plug in the given y and x point than it would turn into $$ 10 = a(5^2 - 6(5) + 5)$$ $$ 10 = a(25 - 30 + 5)$$ $$ 10= a(0)?$$ Is this how you find a? How would you do this?
Our parabola has at most two zeros, that we know, so it is $p(x) = a(x-1)(x-5)$. Now put in the last point you get $p(0) = a(-1)(-5) = 10$ thus $a=2$.
$p(x) = 2(x-1)(x-5) = 2x^2 -12x + 10$.