determine the equation of the quadratic function represented by the table of values below: x -4 -3 -2 -1 0 1 y -20 -13 -8 -5 -4 -5

y = -1*x^2 - 4

Step-by-step explanation:

A generic quadratic equation is written as:

y = a*x^2 + b*x + c

we need to find the values of a, b, and c

We have the table:

x     y = f(x)

-4     -20

-3       -13

-2        -8

-1         -5

0         -4

1          -5

In the table we can see two things:

y(0) = a*0^2 + b*0 + c = c = -4

Then we know that:

y(x) = a*x^2 + b*x - 4

Now we only need to find a and b.

We also can see that we have symmetry around the point x = 0, this means that x = 0 is the vertex of the quadratic equation, and for a general case the vertex is at:

x = -b/2a

and this is equal to zero:

-b/2a = 0

then we must have b = 0.

So for now, the equation is:

y = a*x^2 -4

Now we can just replace the values of any of the points to find the value of a, for example we can use the point (1, -5)

this means that:

y = -5 = a*1^2 - 4

    -5 = a - 4

    -5 + 4 = a

        -1 = a

Then the quadratic equation is:

y = -1*x^2 - 4

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

answer:

f(5/2) = 10/3

i hope this you!

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v