# transformation of functions question: find the range of f

#### bumblebee123

##### Junior Member
Can anyone explain the answer to this question? I think I'm missing something, or here's there's a rule I've forgotten.

question: the function f is defined for all values of x as f(x) = 2x^2 - 10x - 5

a) write 2x^2 - 10x - 5 in the form a(x + b)^2 + c

I managed to do this, the answer is 2 ( x - 5/2 )^2 - 35/4

b) Find the range of F

This is where I get stuck, I have no idea ( or maybe just can't remember ) what to do next!

any help would be really appreciated #### MarkFL

##### Super Moderator
Staff member
If you know where the vertex of a parabolic function is, and which direction the parabola opens (either up or down), then you can use this information to find the range of the function. If expressed in the form:

$$\displaystyle f(x)=a(x-h)^2+k$$ where $$a\ne0$$

Then if $$a<0$$ the range is $$\displaystyle (-\infty,k]$$ and if $$0<a$$ the range is $$\displaystyle [k,\infty)$$.

Does that make sense?

• GetThroughDiffEq

#### bumblebee123

##### Junior Member
If you know where the vertex of a parabolic function is, and which direction the parabola opens (either up or down), then you can use this information to find the range of the function. If expressed in the form:

$$\displaystyle f(x)=a(x-h)^2+k$$ where $$a\ne0$$

Then if $$a<0$$ the range is $$\displaystyle (-\infty,k]$$ and if $$0<a$$ the range is $$\displaystyle [k,\infty)$$.

Does that make sense?
Sorry, I really don't understand this. I don't think my math teacher has taught this to me

#### Dr.Peterson

##### Elite Member
If you haven't seen the vertex form of a parabola, then just use the fact that (x - 5/2)^2 is always positive or zero, and is zero when x - 5/2 is zero. So x = 5/2 gives the lowest possible value of the expression, and that lowest value is the constant term, -35/4.

It will really help us help you if you tell us what you HAVE learned, so we can use that to explain new things. As it is, we can only use what we THINK you should be familiar with, not knowing what course you are taking. Or, if you were given any examples like this, you could show them to us so we can see what method you are expected to use. (If you can't remember what you were taught, then looking back at your book or notes is a good idea anyway! Learning math is not a closed-book test.)

#### bumblebee123

##### Junior Member
If you haven't seen the vertex form of a parabola, then just use the fact that (x - 5/2)^2 is always positive or zero, and is zero when x - 5/2 is zero. So x = 5/2 gives the lowest possible value of the expression, and that lowest value is the constant term, -35/4.

It will really help us help you if you tell us what you HAVE learned, so we can use that to explain new things. As it is, we can only use what we THINK you should be familiar with, not knowing what course you are taking. Or, if you were given any examples like this, you could show them to us so we can see what method you are expected to use. (If you can't remember what you were taught, then looking back at your book or notes is a good idea anyway! Learning math is not a closed-book test.)
I'm doing GCSE maths- I've learnt about functions, inverse functions, composite functions and I think I know about the basics of the vertex form of a parabola. But this type of question has only appeared in a practice book for GCSE maths so I'm a bit confused. I've been reading notes ( which hasn't helped as they've never taught me how to handle a question with lots of different things in like this ) and also watching videos. I kept reading over what you guys wrote but I think it's too advanced for my brain to understand at the moment haha. So using my knowledge from math videos and what I've been taught, I've attempted to understand the question:

2x^2 - 10x -5 is going to be a parabola that's shaped like a bucket, with a minimum point. I know this because the x^2 coefficient is positive.

so the y-coordinates for the parabola ≥ 0

so f(x) must be ≥ 0 as well

f(x) =2x^2 - 10x -5 which is the domain

the minimum point of the parabola is when 2x^2 - 10x -5 = 0

so when 2 ( x - 5/2 )^2 - 35/4 = 0

x - 5/2 = 0
x = 5/2

if i put this back into the equation to find the y-coordinate ( the range? ): 2( 5/2 -5/2 )^2 - 35/2
2 ( 0 )^2 - 35/2
- 35/2
so for the range f(x) ≥ -35/2

I'm not really sure I completely understand what I did, as its mostly copying what I've seen. Is this the correct method though?

#### Dr.Peterson

##### Elite Member
2x^2 - 10x -5 is going to be a parabola that's shaped like a bucket, with a minimum point. I know this because the x^2 coefficient is positive.

so the y-coordinates for the parabola ≥ 0

so f(x) must be ≥ 0 as well
Yes, it opens upward, so it has a minimum. But that doesn't mean that y = f(x) is always at least 0. It means, rather, that y is always at least the minimum, whatever that is! My guess is that you meant something like that, but said something else.
f(x) =2x^2 - 10x -5 which is the domain
Huh? That's just the function. Its domain is the set of all x for which f(x) is defined -- which, in the case of any polynomial, is all (real) numbers.
the minimum point of the parabola is when 2x^2 - 10x -5 = 0

so when 2 ( x - 5/2 )^2 - 35/4 = 0
No. Solving this gives the x-intercepts, where the graph crosses the x-axis because y = 0.
x - 5/2 = 0
x = 5/2
This is the location of the vertex, but it does not come from solving the equation you wrote. It can be found by solving, instead, 2 ( x - 5/2 )^2 = 0.
if i put this back into the equation to find the y-coordinate ( the range? ): 2( 5/2 -5/2 )^2 - 35/2
2 ( 0 )^2 - 35/2
- 35/2
so for the range f(x) ≥ -35/2
Yes, that is correct.

#### Otis

##### Senior Member
Hello bumblebee123. Here's a handy formula to remember, when working with quadratic equations and their graphs.

Given the form y = Ax^2 + Bx + C, the x-coordinate of the vertex point is always -B/(2A).

In your function, B = -10 and A = 2. Therefore:

x-coordinate of vertex = -B/(2A) = 10/(2*2) = 5/2

This formula works with all parabolas that open upward or downward.

As you know, once you have the x-coordinate of a point, you find the corresponding y-coordinate using the given function:

y-coordinate of vertex = 2(5/2)^2 - 10(5/2) - 5 = -35/2

You also know the parabola opens upward, so the vertex point (5/2,-35/2) is the lowest point on the graph. In other words, -35/2 is the smallest function output, so the range must be all Real numbers that are -35/2 or larger.

That fact is what MarkFL discussed, in post #2.

PS: If you're allowed to use graphing software, then work alongside a graph to confirm your thoughts as you go. (That will help prevent mistakes like thinking y cannot be negative in this exercise.) Cheers #### Jomo

##### Elite Member
Sorry, I really don't understand this. I don't think my math teacher has taught this to me
Your teacher can't show everything. You have to think as well. If you know the vertex you should be able to know the range. The vertex is either the lowest point or the highest point.

• bumblebee123

#### bumblebee123

##### Junior Member
I think I understand a bit more now- I'll try it again, hopefully with the right methods this time!

2x^2 - 10x -5 is a parabola with a minimum point

to find the location of the vertex, I need to find the x-intercept.

2( x - 5/2 )^2 = 0
x - 5/2 = 0
x = 5/2

with the x-coordinate of the vertex, I can find the y-coordinate of the vertex ( the range )

y = 2 (5/2 - 5/2 )^2 - 35/2

y = - 35/2

range: f(x) ≥ -35/2

have I done it right? #### HallsofIvy

##### Elite Member
Yes, that is correct. The point of "completing the square" is that a square is never negative! Once you a quadratic written as "(x- a)^2+ b" you know it is "b plus some non-negative number" so it value cannot be lower than b.

#### bumblebee123

##### Junior Member
I think I also could have done this by knowing that y = ( x - h )^2 + k

and the vertex coordinates will be (h, k )