Question 716254
When a number is a root of a polynomial function then (x - that number) is a factor of that function. So the quadratic function you are looking for, in factored form, will be:
{{{f(x) = a(x-(-8))(x-(-6))}}}
or
{{{f(x) = a(x+8)(x+6)}}}
In general, the "a" can be any non-zero number. But the problem states that "a" should be less than zero (or negative) and that we must have a vertex at (-7, 2). So there will probably be only one value for "a" that will fit. First we will multiply out the factors. Using FOIL on the last two factors:
{{{f(x) = a(x*x+x*6+8*x+8*6)}}}
{{{f(x) = a(x^2+6x+8x+48)}}}
{{{f(x) = a(x^2+14x+48)}}}
Distributing the "a":
{{{f(x) = ax^2+14ax+48a)}}}<br>
Now we set out to figure out what "a" must be. The vertex, (-7, 2), should fit this equation and we can use this to find "a". Substituting -7 for x and 2 for y/f(x) we get:
{{{2 = a(-7)^2+14a(-7)+48a)}}}
Now we solve for "a". We start by simplifying...
{{{2 = a(49)+14a(-7)+48a)}}}
{{{2 = 49a+(-98a)+48a)}}}
{{{2 = -a)}}}
Dividing by -1:
{{{-2 = a}}}