Question Next, the team wants to explore how it can change the steepness of the curved pit.
Identify how the graph of each equation compares with the graph of the parent quadratic
equation, y = x.
Drag the equations to the correct location on the chart. Not all equations will be used.
2
Steeper than y = x
Loss Steep than y = x?

Steeper: y = 2x², y = (2x)²

Less Steeper: y = 1/2x², y = (1/2x)²

Step-by-step explanation:

Steeper: If The coefficent of x² is greater than 1, the parent function will be vertically stretched and will be more steeper

Less Steeper: If the coefficent is between 0 and 1, graph of the parent function will be vertically compressed and will be less steeper.

Hope this helps!

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let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!