How many x-intercepts does y= 3x² - 5x + 2 have?

Question

can you show your work if possible........?

Answer 1

To get the x intercept, you have to set y = 0

y= 3x² - 5x + 2
0 = 3x² - 5x + 2

Solve for the x (quadratic equation)
3x² - 5x + 2 = 0
x = (-b ± √[b² - 4ac])/2a
where:
a = 3
b = -5
c = 2
x = {-(-5) ± √[(-5)² - 4(3)(2)]}/2(3)
x = {5 ± √[25 - 24]}/6
x = {5 ± √1}/6
x = (5 + 1)/6 = 1
x = (5 - 1)/6 = 2/3

x = 1, 2/3 <<< these are your 2 x-intercepts

Answer 2

2

Answer 3

A polynomial has as many x-intercepts as the degree of the polynomial, but that includes real and imaginary. To see what type of intercepts you will have use the discriminant b^2-4ac. If you get a positive answer, you have two real x-intercepts. If you get a zero answer, you have one intercept with a multiplicity of two. If you get a negative answer, you will have two imaginary x-intercepts that are complex conjugates.

b^2-4ac=25-24=1

So you have two real x intercepts. Which are:

(5+-1)/6 = 2/3 and 1

Answer 4

y= 3x² - 5x + 2
for there to be x intercepts y should be 0
generally in quadratic equation there will be 2 intercepts

0= 3x² - 5x + 2
0= 3x² - 3x-2x + 2
0= 3x(x-1)-2(x - 1)
0= (3x-2)(x-1)

so
3x-2=0
3x=2
x=2/3

or

x-1=0
x=1

so x either equals 2/3 or 1
there r 2 intercepts

Answer 5

It has 2 real roots. Roots, or solutions = x-intercepts.

The discriminant, which is b^2-4ac, tells how many roots and what type of roots the equation has. If the discriminant is greater than 0, there are 2 real roots. If equal to 0, 1 real root. And if less than 0, 2 imaginary roots.

As for finding the roots, you can use the quadratic formula, which is:

-b[+or-] the square root of b^2-4ac / 2a

In your problem, where a=3, b=-5, and c=2, there would be two roots, being: 5 + the square root of 1 / 6, and 5 minus the square root of 1 / 6

Keep in note that all of -b [+or-] the square root of the b^2-4ac (the discriminant) is -all- over 2a.

Answer 6

this is a rule:
plug it into the quadratic formula and look at the square root part(discriminant or something like that).
if its less than 0 (negative) then no x-intercepts
if it IS 0 then 1 x-intercept
if above any # above 0, 2 x-intercepts.

square root of (-5)^2-4(3)(2)
square root of 25-24
square root of 1
meaning 2 x-intercepts because 1 is greater than 0

to factor:
its (3x-2)(x-1)
x = 2/3 and 1

Answer 7

it has 2 intercepts, you have to factor the problem i got
3x(x-o)2(x+1) so the 2 intercepts are 0 and -1, hope i helped!

Answer 8

For positive large x, y is also large. And for negative large X, y is also large. Somewhere in between, y is at its minimum value. At this point, the derivative is zero:

6x - 5 = 0

So 6x = 5, or x = 5/6
Plug this back into the equation for y, we get y = -0.08333...
For values of x away from x = 5/6, the y value will be larger than this. If you go in the negative direction, the graph will hit y = 0 for some value of X. And if you go in the positive direction, the graph will hit y = 0 for some value of X.

So there are two x-intercepts, one less than 5/6, and one larger than 5/6. They are at:

x = 0.666..., y = 0, and
x = 1.000, y = 0

Answer 9

y=3x^2-5x+2
first factor. : (3x-2)(x-1)
to find x int. replace y=0
3x-2=0 [bring 2 on the right side and change the sign]
3x=0+2 [then divide 3x on both sides]
thus, x= 2/3

the second bracket: do the same. replace y=0
thus, x-1=0
x= 1.
thus::::: there are 2 x-inter.{{{{ x=2/3 and x=1}}}}

I think this is how it's done.

Answer 10

y=3x² - 5x + 2
Set y=0, then factor.
0=(x-1)(3x-2)
Since there are two possible answers for x (x=1 or x = 2/3), there are 2 x-intercepts.