Question

Create a quadratic function f(x), other than f(x) = x^2, and write it below. Then below the function, state the domain, range, x-intercepts, y-intercept, and f(3).

Answer 1

`f(x)=2x^(2)-3`

`D=(-\infty,\infty)`

`R=[-3,\infty)`

x-intercepts:
`((√(6))/(2),0)` and
`(-(√(6))/(2),0)`

y-intercept: (0,-3)

f(3)=15

Explanation:

A quadratic function is any polynomial function whose greatest power is a 2, so
`f(x)=2x^(2)-3` is a nice example of it.

For the domain, as a rule, polynomial functions that have positive whole degrees will be defined for all real numbers, so:

`D=(-\infty,\infty)`

When talking about quadratic functions, its range will happen from the lowest point (in this case the vertex) to it's highest point (in this case infinity)

In order to find the vertex, we can use the following formula:

`x=-(b)/(2a)`

which is used for quadratic equations in standard form:

`y=ax^(2)+bx+c`

In this case a=2, b=0 and c=-3, so we get:

`x=-(0)/(2(2))=0`

Now we know that the range will happen at x=0, so we substitute:

`f(0)=2(0)^(2)-3=-3`

so the range will be:

`R=[-3,\infty)`

For the x-intercepts we need to set the given function equal to zero so we get:

`2x^2-3=0`

and solve for x:

`2x^(2)=3`

`x^(2)=(3)/(2)`

`x=\pm \sqrt{(3)/(2)}`

which can be rationalized to:

`x=(√(6))/(2)` and
`x=-(√(6))/(2)`

so the x-intercepts are located at:

`((√(6))/(2),0)` and
`(-(√(6))/(2),0)`

For the y-intercept, we need to set x=0 so we get:

`f(0)=2(0)^(2)-3`

`f(0)=-3`

So the y-intercept is located at:

(0,-3)

and finally, in order to get f(3) we must substitute x for 3 so we get:

`f(3)=2(3)^(2)-3`

f(3)=15