Find two quadratic functions

Question

Find two quadratic functions, one that opens upward and one that opens downward, whose graphs have the given x-intercepts. (There are many correct answers.)

(1, 0), (3, 0)

Answer 1

Hi Tatyana

f(x) = x² -4x + 3 this one opens upward

For the Standard Form below, which can be the result of binomial multiplication, that typically can be un FOILED based on factors of the constant c and x with b being the coefficient of the sum of first degree products

f(x) = ax² + bx + c

Since you have coordinates for the x intercepts from a factored quadratic where the y coordinate, or f(x) =0

You have (1, 0) and (3, 0) and order for x to be positive the factors of c must be negative

You can plug them in

0 = (x -1)(x - 3)

A quick check will confirm the x intercepts

x - 1 = 0

x =1

and

x -3 = 0

x =3

To get your first quadratic function just FOIL your binomials

f(x) = (x -1)(x - 3)

f(x) = x² -3x -x + 3

Combine your terms

f(x) = x² -4x + 3 there is one quadratic

Of course you can graph this to confirm the given x intercepts and also the y intercept (0, 3)

You can use the info above or similar data, for example the vertex form to find another give it a try

Using the vertex form f(x) = a(x - h)² + k.

I got

f(x) = -(x - 2)² + 1

Notice in the vertex form above, a = -1 this point the parabola downward

I hope you find this useful if you have any questions send me an email.