Suppose $f(x)$ is a quadratic function such that $f(1)=-24,$ $f(4)=0,$ and $f(7)=60$. determine the value of $f(-1)$.suppose $f(x)$ is a quadratic function such that $f(1)=-24,$ $f(4)=0,$ and $f(7)=60$. determine the value of $f(-1)$.

-20

Step-by-step explanation:

Well we are given that is a quadratic equation.

We are also given f(4) = 0. This means 4 is one root of the quadratic.

We can factor the quadratic into a form like this:

f(x) = a (x - r) (x - 4)

Where r is the other root of the quadratic.

We can subsitute 1 and 7 into the quadratic because we are given that

f(1) = - 24 and f(7) = 60.

We will get the following:

-24 = a (1 - r) (-3)

and

60 = a (7 - 4) (3)

To get rid of the "a" term, we can divide the second equation by the first, giving us the following equations:

- 5/2 = - (7 - r) / (1 - r)

We can multiply both sides by 2 (1 - r).

This gives us:

-5 ( 1 - r) = -2 (7 - r), so

5r - 5 = 2r - 14, so

3r = -9.

Then, we know r =  -3.

We can then derive that the unique quadratic with the desired values is:

f(x) = 2 (x + 3) (x - 4).

We can plug in x = -1, so we get f(-1) = (2) (2) (-5) = -20.

Thus, our answer is -20.

To be honest, the first way or the other soultion the one I thought of first. In the middle of doing this problem for fun, I discovered this simpler way.

The value of f(-1) is:

Let the quadratic function f(x) be denoted by:

Now, it is given that:

This means that:

Also,

This means that:

on subtracting equation (1) from equation (2) we get:

Also,

i.e.

On subtracting equation (1) from equation (3) we have:

on subtracting equation(x) from equation(y) we have:

on putting the value of 'a' back in equation (x) we have:

Also, on putting the value of a and b in equation (1) we have:

Hence, the quadratic function is:

Hence,