quadratic function(graphing help)

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perilous

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Oct 18, 2005
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Okay, I'm giving you the problem, I know how to solve most of it, just not sure if I did it correctly :?
Sketh the graph of the quadratic function y=3x^2+3x-2
the vertex: (.21,-1)
the x intercepts-1.4,.45)
the y intercept -2
Are those right?

Okay, I'm graphing it. You use your maximum to get the vertex, but the graph on my calc. shows the vertex being on the - side, and I have .21?

Some other issues I'm having with it, are the second part of the problem:
3x^2+3x-2=0
" " greater then or equal to 0
" " less then 0

Thanks for any help, my teacher goes so fast, I think I only get half of what she says, then I'm always lost.
 
Okay, I'm giving you the problem, I know how to solve most of it, just not sure if I did it correctly Confused
Sketh the graph of the quadratic function y=3x^2+3x-2
the vertex: (.21,-1)
the x intercepts-1.4,.45)
the y intercept -2
Are those right?

Okay, I'm graphing it. You use your maximum to get the vertex, but the graph on my calc. shows the vertex being on the - side, and I have .21?
Click to expand...
There are several approaches to find the vertex.

You mention "you maximum" which hints calculus. The calculus approach would be to differentiate y=3x^2+3x-2 to get dy/dx = 6x + 3. Then to set it to zero:
0 = 6x + 3
6x = -3
x = -0.5

So at the vertex, y = 3(-0.5)^2+3(-0.5)-2 = -2.75

So the parabola is centred, ie. has a vertex at, (-0.5, -2.75).

One algebraic approach:

y = 3x^2+3x-2

y = 3[x^2 + x] - 2
y = 3[ (x+0.5)^2 - 0.25] - 2 <- complete the square
y = 3(x+0.5)^2-0.75 - 2
y = 3(x+0.5)^2 - -2.75

So the parabola is centred on (-0.5, -2.75)

Another algebraic approach:

The roots of the quadratic can be found using the quadratic formula, and these are the x-intercepts:

x = [-b +/- sqrt(b^2 - 4ac)]/2a
x = [-3 +/- sqrt(9 - 4*3*(-2)]/6
x = 0.457, -1.46 (3sf)

That is, you're correct with your x-intercepts (if we ignore rounding mistakes).

The x-ordinate of the vertex will be exactly half-way between these two points. **

x = (0.457 + (-1.46))/2 = -0.500 (3sf)

This agrees with the above methods.

The y-intercepts occur when x = 0. You are absolutely correct.

In other words, you got everything but the vertex correct. Therefore the only usefulness of this post (if at all) will come from **.
 
Some other issues I'm having with it, are the second part of the problem:
3x^2 + 3x-2 = 0
3x^2 + 3x - 2 >
3x^2 + 3x - 2 < 0
Click to expand...
I have to go to school but quickly...

You can use the information you have obtained (vertex, x-intercepts, y-intercept) to draw the graph. You can then see in which intervals the graph is above or below y=0.

See that it is below y=0 (ie. below the x-axis) between the x-intercepts, and is above the y=0 for all other x-values.
 
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