Wich inequality statement best represents the graph ? serious answers only!

a) f(x) ≤ -2x^2 + 4x

Step-by-step explanation:

In the graph you can see that the "hands" of the quadratic function go up, this means that the principal constant (the one that multiplicates the square term) is positive:

this is:

for a function f(x) = a*x^2+ b*x + c

if a is positive, the hands will go upside, if a is negative, the hands will go downside.

we also can see that the shaded area is upside the curve, so we are looking for f(x) > something.

Then we can discard the third option because we have f > -2*x^2..

the remaining options are:

a) f(x) ≤ -2x^2 + 4x or f(x) ≥ 2x^2 - 4x

b) f(x) ≥ 2x^2 + 4x

first, you can see that in the second function both terms have the same sign, so there can be only one x-intercept, but in the graph, we can see two x-intercepts, so we can also discard option b.

The remaining option, and the correct one, is option a.

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!

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step-by-step explanation:

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