Find a Quadratic polynomial function with real coefficients satisfying the given conditions.
-4 and 3 are zeros; f(1) = -30
HELP. I have no idea.
since the zeros of the quadratic are -4 and 3
we can say:
f(x) = a(x+4)(x-3)
so the only thing we don't know is a possible coefficient.
but f(1) = - 30
-30 = a(1+4)(1-3)
-30 = -10a
a = 3
f(x) = 3(x+4)(x-3)
expand if so needed
To find a quadratic polynomial function with real coefficients that satisfies the given conditions, we can use the fact that if a quadratic polynomial has zeros at x=a and x=b, then it can be written in the factored form as follows:
f(x) = a(x - r)(x - s),
where r and s are the zeros of the polynomial.
In this case, the zeros are -4 and 3, so we can write the quadratic polynomial as:
f(x) = a(x + 4)(x - 3).
Now, to find the value of 'a', we can use the additional condition that f(1) = -30. This means that when x = 1, the value of f(x) (or y) is -30. We can substitute these values into the quadratic equation to solve for 'a'.
Let's evaluate the function for x = 1:
f(1) = a(1 + 4)(1 - 3) = a(5)(-2) = -10a.
We know that f(1) = -30, so we can set up the equation:
-10a = -30.
Now, solve for 'a':
Divide both sides by -10: a = -30 / -10 = 3.
Therefore, the quadratic polynomial that satisfies the given conditions is:
f(x) = 3(x + 4)(x - 3).