Let y=2x^-5x+2. Find all Intercepts?

Question

Please be detailed

Answer 1

Y-intercepts are when x=0
then from y=2x^-5x+2 at x=0 y=2 +>intercept point (0,1)

X- intercepts when y=0
then 2x^-5x+2=0 solve for x
using form ax^2+bx+c=0
x1=(-b+sqrt(b^2-4ac))/(2a)
x2=(-b-sqrt(b^2-4ac))/(2a)
x1=(5+sqrt(25-4(2)2))/4=
x1=(5+sqrt(9)/4=
x1=(8/4)=2
x2=(5-3)/4=1/2
points for x-intercept are (1/2.0) and (2,0)

PS Carlos G provided you with a good reference

Answer 2

Is the equation y = 2x^2 - 5x + 2 ?

If so, first of all, let y = 0, and solve for x.
2x^2 - 5x + 2 = 0
(2x-1) (x-2) = 0
2x - 1 = 0 (or) x - 2 = 0
x = 1/2 (or) x = 2
x intercepts: {1/2, 2}

Now let x = 0, and slove for y.
y = 2(0)^2 - 5(0) + 2 = 2
y intercpet: {2}

Answer 3

y = 2x^2 -5x + 2

x-intercepts
Set y = 0 and solve for x
y = 0 = 2x^2 -5x + 2
(2x - 1)(x - 2) = 0 ----> by trial and error, or use quadratic formula
(2x - 1) = 0 or (x - 2) = 0
x = (1/2) or 2

x-intercepts are: ( 1/2 , 0) and (2 , 0)


y-intercepts
Set x = 0 and solve for y
y = 2(0)^2 -5(0) + 2
y = 2

y-intercepts is: (0 , 2)

Answer 4

y = 2x^2 - 5x + 2

x = (-b ± sqrt(b^2 - 4ac))/2a

x = (-(-5) ± sqrt((-5)^2 - 4(2)(2)))/(2(2))
x = (5 ± sqrt(25 - 16))/4
x = (5 ± sqrt(9))/4
x = (5 ± 3)/4
x = (2/4) or (8/4)
x = (1/2) or 2

Intercepts : (2,0), ((1/2),0), and (0,2)

Answer 5

Summer school homework?

Learn the topic here: http://purplemath.com/modules/intrcept.htm

Answer 6

*sniff* ....*sniff*....do I smell a homework question.......*sniff* yep.


Sorry....not going to help you cheat. Figure it out yourself.