Preliminary Quadratics Question (1 Viewer)

[Direwolf]

Could someone please explain to me how to do this question?

Determine the greatest value of the function f(x) = 17 + 4x - x^2

Appreciate the help

 

[Peeik]

Hint: The greatest value is at the vertex because this is a concave down parabola. So how do you find the vertex of a parabola?? If you know this, then it will be your answer.

 

[deswa1]

Could someone please explain to me how to do this question?

Determine the greatest value of the function f(x) = 17 + 4x - x^2

Appreciate the help

There are a few ways to do this. I don't know if you know calculus yet (I'm assuming you would) but I'll do it without calculus just it case:

f(x)= 17+4x-x^2
f(x)= -(x^2-4x)+17
f(x)=-(x^2-4x+4)+17+4
f(x)= -(x-2)^2 +21

Now notice that any number squared must be positive. Therefore the smallest possible value of (x-2)^2 is 0 when x=2.

Therefore the maximum value of f(x)=21 when x=2

 

[Direwolf]

There are a few ways to do this. I don't know if you know calculus yet (I'm assuming you would) but I'll do it without calculus just it case:

f(x)= 17+4x-x^2
f(x)= -(x^2-4x)+17
f(x)=-(x^2-4x+4)+17+4
f(x)= -(x-2)^2 +21

Now notice that any number squared must be positive. Therefore the smallest possible value of (x-2)^2 is 0 when x=2.

Therefore the maximum value of f(x)=21 when x=2

I was trying to do it this way, thanks a lot. We haven't learnt calculus yet ):

 

[Leffife]

Determine the greatest value of the function f(x) = 17 + 4x - x^2

f'(x) = 4 - 2x
f''(x) = - 2

Find stationary points by f'(x) = 0

4 - 2x = 0
2x = 4
x = 2 , y = 15

f''(x) = -2 < 0

Thus (2,15) is maximum turning point.

 

[deswa1]

Determine the greatest value of the function f(x) = 17 + 4x - x^2

f'(x) = 4 - 2x
f''(x) = - 2

Find stationary points by f'(x) = 0

4 - 2x = 0
2x = 4
x = 2 , y = 15

f''(x) = -2 < 0

Thus (2,15) is maximum turning point.

When x=2, y=21

 

[Leffife]

When x=2, y=21

Yup lol what he said, I realised I substituted a different value instead of x = 2 into it >.<