# Sketch graph of function f(x) = 2(x-1)^2 - 3

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• February 7th 2010, 07:51 AM
shadow6
Sketch graph of function f(x) = 2(x-1)^2 - 3
The question states: Determine the domain and range of the function f(x) = 2(x-1)^2 - 3 by sketching its graph.

I know how to determine D and R, though I forgot how to sketch parabolas on graph. I think I also forgot how to factor too...

So far I did the following:

f(x) = 2(x - 1)^2 - 3
0 = 2(x - 1)^2 - 3
0 = 2(x - 1)(x - 1) - 3
0 = 2x^2 - 1x -1x + 2 - 3
0 = 2x^2 - 2x - 1

What do I do from here? Is this correct so far, or did I mess up on the first line (with the 2)?
• February 7th 2010, 08:10 AM
felper
$f(x) = 2(x-1)^2 - 3$

At first, note that the coefficient that multiplies to x^2, is postive. With this information, you know that the parabola opens as a smile. Now, where's the vertex? It's on (1,-3) (can you tell us why?). Now, you can sketch the graph of the function. You'll know that the function takes all the values greater or equal to the vertex. Which would be the range?
• February 7th 2010, 08:25 AM
shadow6
http://www.mathhelpforum.com/math-he...73cd2adb-1.gif

"At first, note that the coefficient that multiplies to x^2, is postive. With this information, you know that the parabola opens as a smile."

Had this already; a positive 2 = minimum value.

"Now, where's the vertex? It's on (1,-3) (can you tell us why?)."

This is where I got confused. The -3 is the y value, though I wasn't sure as to how to get the x-value. My guess is that the brackets must = 0? Therefore 1 - 1 = 0?

Now, you can sketch the graph of the function. You'll know that the function takes all the values greater or equal to the vertex. Which would be the range?

Think I got it. Because the two x intercepts can be anywhere on the x axis, then the domain = {xER}. And because the parabola opens where y = -3, any y coordinate below -3 is not part of the graph, making the range = {ER|y>=-3}.

This look correct? (Just need clarification on the x value = 1.)
• February 7th 2010, 08:29 AM
felper
Yes, your right with the range!

The x is not restricted by anything, and the y can't reach values below -3. To solve the confussions you have, read this article

Parabola - Wikipedia, the free encyclopedia
• February 7th 2010, 08:32 AM
shadow6
Thx.
• February 7th 2010, 10:08 AM
shadow6
Another question is asking:

Determine the range if the domain is {-3, -1, 0, 2.5, 6}

a) f(x) = 4-3(x) b) f(x) = 2x^2 - 2x + 1

Do I just sub the x values into the equation to solve for the y values?
• February 7th 2010, 10:11 AM
Raoh
Quote:

Originally Posted by shadow6
Another question is asking:

Determine the range if the domain is {-3, -1, 0, 2.5, 6}

a) f(x) = 4-3(x) b) f(x) = 2x^2 - 2x + 1

Do I just sub the x values into the equation to solve for the y values?

hello
the range would be, $\left \{ {f(-3), f(-1), f(0), f(2.5), f(6)} \right \}$.
• February 7th 2010, 10:14 AM
shadow6
For which question, a or b? And how do you get the range? Are they they same as the domain or do you have to sub something?
• February 7th 2010, 10:18 AM
Raoh
Quote:

Originally Posted by shadow6
For which question, a or b? And how do you get the range? Are they they same as the domain or do you have to sub something?

for any function defined on those points.
• February 7th 2010, 10:20 AM
shadow6
Alright thx.
• February 7th 2010, 02:09 PM
shadow6
Another question gives me an equation f(x) = -2(x + 1)^2 + 3

Then it tells me to determine:

i) f(1) - f(0)

ii) 3f(2) - 5

iii) f(2 - x)

I started with the first.

I subbed f(1) - f(0) into f(x) = -2(x + 1)^2 + 3.

f(x) = -2(x + 1)^2 + 3
f(1) - f(0) = -2((1 - 0) + 1)^2 + 3
f(1) - f(0) = -2(1 + 1)^2 + 3
f(1) - f(0) = -2(2)^2 + 3
f(1) - f(0) = -2(4) + 3
f(1) - f(0) = -8 + 3
f(1) - f(0) = -5

That look right? Did I sub correctly?
• February 8th 2010, 04:04 AM
Raoh
Quote:

Originally Posted by shadow6
Another question gives me an equation f(x) = -2(x + 1)^2 + 3

Then it tells me to determine:

i) f(1) - f(0)

ii) 3f(2) - 5

iii) f(2 - x)

I started with the first.

I subbed f(1) - f(0) into f(x) = -2(x + 1)^2 + 3.

f(x) = -2(x + 1)^2 + 3
f(1) - f(0) = -2((1 - 0) + 1)^2 + 3
f(1) - f(0) = -2(1 + 1)^2 + 3
f(1) - f(0) = -2(2)^2 + 3
f(1) - f(0) = -2(4) + 3
f(1) - f(0) = -8 + 3
f(1) - f(0) = -5

That look right? Did I sub correctly?

it doesn't look right.
$f(1)=-5$.
$f(0)=1$.
$f(1)-f(0)=-5-1=-6$.