How do you find the x-intercepts for the parabola: y= 3x^2 + 2x + 1?

Answer 1

To find the X-intercept put y=0
Therefore
3x^2 + 2x +1=0
solve for x we find that there is no real values of x, so there is no x-intercept. This parabola is opening up and is above the x-axis.
To find the vertex
3x^2 + 2x +1=y
divide by 3 to make the coefficent of x equal to 1
x^2 + 2/3 x +1/3=y
x^2 + 2(1/3) x +1/3=y
Add (1/3)^2 and subract (1/3)^2

{x^2 + 2(1/3) x +(1/3)^2}+{-(1/3)^2+1/3}=y
(x + !/3)^2 + (1/3 - 1/9) =y
(x + !/3)^2 + (2/9) = y
The coordinate of vertex are x = (-1/3) and y = sqrt of (2/9) which are above the x-axis therefore the parabola will not intercept the x-axis

Answer 2

TC is right, but I'll explain how to determine if there are any x-intercepts. When solving the quadratic equation y(x)=0, look for its discriminant
D=b^2-4*a*c for a^2+b*x+c=0.
If D>0 there are 2 x-intercepts and 2 different roots.
If D=0 there is one x-intercept and 2 roots wich are the same.
Finally, when D<0 there are no roots(I mean roots which belong to the Real set only, of course) and no x-intercepts as you guess =)
In your case D=2^2-4*3*1=4-12=-8<0
The answer is: there are no x-intercepts

Answer 3

the vertex is the area the place the curve 'bends' to locate the coordinates of the vertex you're able to desire to end the sq. for 3x^2 + 6x - 12 it is putting it interior the type a(x - b)^2 + c the place (b,c) is the vertex the working... 3x^2 + 6x - 12 = 3(x^2 + 2x) - 12 = 3((x + a million)^2 - a million) - 12 = 3(x + a million)^2 - 3 - 12 = 3(x + a million)^2 - 15 so the vertex of the parabola is at (-a million,-15) it is the backside factor on the curve because it slopes downwards then upwards. desire it is clever.

Answer 4

b^2 -4ac should be more that 0, in order for the equation to have x-intercepts!
while in your equation: (2^2) -4*3*1 = 4 -12 = -8
-8 is less than zero and thus there are no real roots (no x-intercepts) for the equation, unless which i doubt that you want some imaginary roots!

Answer 5

to find the x intercepts of anything you want it so y = 0

thus

3x² + 2x + 1 = 0 solve...

you cant factor that easily so use the quadratic

x = -b (+/-) sqrt(b²- 4ac)/2a
= -2 (+/-) sqrt(4 - 12) / 6
= - 2 (+/-) i sqrt(8) / 6
= (- 1 (+/-) i sqrt(2))/3

the roots are imaginary... if you didn't know the quadratic method i really doubt you understand the concept of imaginary numbers... but goodluck

Answer 6

There's actually 3 ways:
Find when y=0 by using the quadratic formula,
Reverse FOIL
and Graph it and see where the parabola touches the x line
Edit: TC is right

Answer 7

Equate y = 0, then solve it using
{-b +- (b^2-4ac)^2 } / 2a

where a, b, c is the symbol of quadratic equation, ax^2 + bx + c

For this case, no real value of x. The curve didnt touch the x-axis.

Answer 8

hmm, looks like this is all above the x axis.
So there aren't any. At least in the real numbers.

Use the quadratic equation and you get
-2 +- sqrt(4-12)/ 6
-2 +-sqrt(-8) / 6
(-2 +- i * 2sqrt(2))/6
1/3 +- i * sqrt(2)/3

Answer 9

since b^2 is<4ac the parbola would not have any x intercepts

Answer 10

equate y=0 Then find
It happens in this case that roots are imaginary
Hence no intercepts
Graph lies above X axis