Solving and Graphing Functions: f(x) = 2x^2 - 3x + 2

Toxxic Kitty

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Sep 9, 2015
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I need help with solving and graphing the following function: f(x) = 2x2 -3x+2

a) Find your exact x-intercepts

b) State your y-intercepts

c) Find your exact vertex

d) Graph f(x) and show and name all of the above information


So far I have the following:

a) Imaginary:

f(x) = 2x2--3x+2 x= [-b +/- √b2-4ac]/2a a= 2 b= -3 c=2

x= [-(-3) +/-√-32-4(2)(2)]/2(2) --> x= [3 +/- √9 - 16]/4 --> x= [3 +/- √-7]/4 --> x= [3 +/- √-1 * √7]/4

x= 3i + √7 /4 , 3i - √7/4



b) (0,2)

f(x)= 2x2-3x+2

f(0) = 2(0)2-3(0)+2 --> f(0) = 0-0+2

f(0) = 2


c) (3/4, 2)

f(x) = 2x2-3x+2 ax2+bx+c a= 2 b= -3 (x,y)= x=(-b/2a) y= f((-b/2a))

x= [-(-3)/2(2)] --> x= (3/4) y= f(3/4) = 2(3/4)2-3(3/4)+2 --> f(3/4)= 9/4-9/4+2 --> f(3/4) = 2


So, now I'm a the graphing part. I know that since my x intercepts are imaginary, that I should only have to plot the vertex. I remember my professor saying that with an equation like this the graph would have two parabolas facing in opposite direction, that do not touch the x axis. I'm trying to figure out how to get the values for the second parabola, and if the work I did for a - c is correct.
 

stapel

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I need help with solving and graphing the following function: f(x) = 2x2 -3x+2

So far I have the following:

a) Find your exact x-intercepts
a) Imaginary:

f(x) = 2x2--3x+2 x= [-b +/- √b2-4ac]/2a a= 2 b= -3 c=2
I'm not sure what's going on with the "minus minus three" in the middle here...? Also, to get all the points, you might want to include all the steps:

. . . . .\(\displaystyle f(x)\, =\, 2x^2\, -\, 3x\, +\, 2\)

. . .find zeroes:

. . . . .\(\displaystyle 2x^2\, -\, 3x\, +\, 2\, =\, 0\)

. . .apply Quadratic Formula:

. . . . .\(\displaystyle x\, =\, \dfrac{-(-3)\, \pm\, \sqrt{\strut (-3)^2\, -\, 4(2)(2)\,}}{2(2)}\, =\, \dfrac{3\, \pm\, \sqrt{\strut 9\, -\, 16\,}}{4}\, =\, \dfrac{3\, \pm\, \sqrt{\strut -7\,}}{4}\)

x= [-(-3) +/-√-32-4(2)(2)]/2(2) --> x= [3 +/- √9 - 16]/4 --> x= [3 +/- √-7]/4 --> x= [3 +/- √-1 * √7]/4

x= 3i + √7 /4 , 3i - √7/4
How did the imaginary (the "i") move from the square root to the leading term?

b) State your y-intercepts
b) (0,2)

f(x)= 2x2-3x+2

f(0) = 2(0)2-3(0)+2 --> f(0) = 0-0+2

f(0) = 2
Yes.

c) Find your exact vertex
c) (3/4, 2)

f(x) = 2x2-3x+2 ax2+bx+c a= 2 b= -3 (x,y)= x=(-b/2a) y= f((-b/2a))

x= [-(-3)/2(2)] --> x= (3/4) y= f(3/4) = 2(3/4)2-3(3/4)+2 --> f(3/4)= 9/4-9/4+2 --> f(3/4) = 2
If you're allowed to use the formula instead of completing the square (in which case, yay!), then check your work on the y-coordinate. Otherwise, do all the steps for completing the square.

d) Graph f(x) and show and name all of the above information
So, now I'm a the graphing part. I know that since my x intercepts are imaginary, that I should only have to plot the vertex.
The vertex, and also the y-intercept.

I remember my professor saying that with an equation like this the graph would have two parabolas facing in opposite direction...
No. If the class had been discussing hyperbolas (and if the equation had been of that type), then you'd have a case. But this is very much NOT true for parabolas, like what you have.

Just plot some extra points, and then rough in the graph. ;)
 

Toxxic Kitty

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How did the imaginary (the "i") move from the square root to the leading term?
Dang. I hadn't even realized I did that.


No. If the class had been discussing hyperbolas (and if the equation had been of that type), then you'd have a case. But this is very much NOT true for parabolas, like what you have.
Okay, so then I'm correct when I only have one parabola on my graph. So I'm just going to plot the vertex and my y-intercept, and that is all right?
 

Toxxic Kitty

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Sep 9, 2015
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You're sure that shouldn't be: 2x^2 - 3x - 2 ?
No I wrote it correctly. The equation on my homework is f(x)= 2x2 -3x +2
 

stapel

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...I'm correct when I only have one parabola on my graph. So I'm just going to plot the vertex and my y-intercept, and that is all right?
No. Even were this a straight line (and it isn't), you'd want at least three plot points. As this line is curved, you'll want plenty of additional plot points. So, as advised earlier, find some additional plot points, and then rough in the graph. ;)
 

Ishuda

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Jul 30, 2014
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...So, now I'm a the graphing part. I know that since my x intercepts are imaginary, that I should only have to plot the vertex. I remember my professor saying that with an equation like this the graph would have two parabolas facing in opposite direction, that do not touch the x axis. I'm trying to figure out how to get the values for the second parabola, and if the work I did for a - c is correct.
How one plots a graph depends on just what one wants to do with the graph and what the instructor requires. Definitely you need the y intercept, (0,2), as a point, the vertex, (0.75,0.875) and, since the parabola is symmetric about the vertex, the mirror point of the y intercept on the other side, (1.5,2). You should label these points. You would also have the x intercepts if they existed on the graph. You now have enough information to sketch a rough graph since you know what a parabola looks like.

However, you probably shouldn't try to read any points off the graph other than the ones you labeled. If you want to use the graph as an estimated of the function value, start filling in the right hand side (or left hand side) and mirror the points. For this problem, I would probably work from the vertex back putting a point midway between the y intercept and vertex and maybe a few more at that spacing going back to the left depending on just how accurate I wanted the graph to be and how much of the graph I wanted.
 
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