Question

For the function defined by: {x^2, x<=1 2x+1, x>1} evaluate f(0) and graph f(x)? Someone help please

Answer 1

The function is defined in two parts:
For all x less than or equal to 1 it is x^2
For all x greater than 1 it is 2x + 1.
To evaluate f(x) for x = 0 first figure out which part to use for x = 0 and substitute 0 for x.

Answer 2

So I would do 0^2, x<=1? That equals 0, but how would I write that as my answer?

Answer 3

f(0) = 0

Answer 4

Oh ok, and how would I graph f(x)?

Answer 5

For all x less than or equal to 1 graph x^2 (which is a parabola)
For all x greater than 1 graph 2x + 1 (which is a straight line)

Answer 6

Do I put x^2 and 2x+1 on the same graph or draw different graphs

Answer 7

Same graph because its is the same function but just has two parts.

Answer 8

Basically, it's just a parabola with a straight line crossing through it

Answer 9

No, the straight line cannot cross the parabola.
Remember the original function.
The parabola is good ONLY for values of x less than or equal to 1.
So the parabola has to stop at x = 1.
The straight line is good only for x > 1.
So the straight line should not go below x less than or equal to 1.

Answer 10

I went to desmos graphing calculator and typed in x^2 and 2x+1. Is there anyway you could show me the graph?

Answer 11

The drawing tool does not provide a proper way to draw a parabola and so I will do a free hand sketch which will likely look awful. But you will get the idea. Give me a few minutes.

Answer 12

Ok, thanks