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

• Feb 7th 2010, 06:51 AM
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)?
• Feb 7th 2010, 07:10 AM
felper
$\displaystyle f(x) = 2(x-1)^2 - 3$

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?
• Feb 7th 2010, 07:25 AM

"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?)."

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.)
• Feb 7th 2010, 07:29 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
• Feb 7th 2010, 07:32 AM
Thx.
• Feb 7th 2010, 09:08 AM

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?
• Feb 7th 2010, 09:11 AM
Raoh
Quote:

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,$\displaystyle \left \{ {f(-3), f(-1), f(0), f(2.5), f(6)} \right \}$.
• Feb 7th 2010, 09:14 AM
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?
• Feb 7th 2010, 09:18 AM
Raoh
Quote:

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.
• Feb 7th 2010, 09:20 AM
Alright thx.
• Feb 7th 2010, 01:09 PM
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?
• Feb 8th 2010, 03:04 AM
Raoh
Quote:

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.
$\displaystyle f(1)=-5$.
$\displaystyle f(0)=1$.
$\displaystyle f(1)-f(0)=-5-1=-6$.