How do you find the x\y intercepts?

Question

My equation is y - 1 = (x - 3)^2
(x minus 3 quantity squared)
Which I changed to x^2 - 6x + 10

the objective is to find the x\y intercepts and the vertex. The vertex i believe is (3,1) providing my math is correct

THE PROBLEM I am having is that I cannot factor the equation and therefore only know of the quadratic equation for solving for the x\y intercepts... Is this correct? The last problem I had like this was x^2 -2x+3 which i plugged into the quadratic equation and came up with 1 +\- i \/'''''''2
(thats i square root of 2) which hardly seems practical as an x or y intercept. Am I doing this right? Thanks for your help

You all have been most helpful THANK YOU ALL SO MUCH

Answer 1

set x=0 to get the y intercept and
y=0 to get the x intercept
in this case
setting x=0
y-1=9
y=10
so y intercept=(0.10)
setting y=0
-1=(x-3)^2
x-3=rt(-1)
x imaginary
so no x intercepts

Answer 2

To find x and y intercepts you must put 0 for the opposite intercept.

For example y=x+3, the y-intercept is 3 because y=0+3. The X-intercept is -3 because 0=x+3 and -3=x

To answer your problem....y - 1 = (x - 3)^2


Y Intercept=
y - 1 = (0-3)^2
y - 1 = 9
y=10

X intercept=
-1 = (x-3)^2
-1 = x^2 - 9
8 = x^2
x= The square root of 8


This is what I believe it is. Then again I am only in freshman Algebra

Answer 3

You won't need to use the quadratic equation for this. You get a y-intercept when x equals zero, so your equation becomes y-1=(-3)^2, which is very easy to solve. Similarly, you get an x-intercept when y equals zero, and your equation becomes -1=(x-3)^2. You can begin solving this one by taking the square root of each side, and you get the square root of a negative number on the left side of the equation, meaning an imaginary number, or no actual x-intercept. If you graph the parabola, you will see that it never crosses the x-axis.

Answer 4

all you gotta do is plug in zero for either y or x in the equation, depending on which intercepts you want to find.

like this: to find the x-intercept(s), plug 0 in for y and solve for x because when y=0, the graph will hit the x-axis:
0 - 1 = (x-3)^2 => -1 = x^2 - 6x + 9 => x^2 - 6x + 10
...after doing this we find that this is not factorable which says there are no x-intercepts.

to find the y-intercept(s), plug 0 in for x and solve for y for the same reason that when x=0, the graph is hitting the y-axis
y - 1 = (0 - 3)^2 => y = 9 + 1 => y-intercept = (0,10)


hope this helped

Answer 5

For the y-intercept, just plug x = 0 into the equation and see what the value of y is. For the x- intercept, you have to solve the equation but a quick check of b^2 - 4ac comes to 36 - 40 = -4 so we know the only roots are complex which means there is no x-intercept, i.e. the graph never crosses the x-axis.

Answer 6

The standard equation in intercept form is x / a + y / b = 1

y- 1 = ( x- 3) ^2
I think question can be like this y -1 = ( x-3 ) / 2
then we can do like this
2 ( y-1) = x-3
2y - 2 = x - 3
2y - x = -3 +2
Multiplying both sides with ( - )
x - 2y = 1
Now X -intercept = 1
Y -intercept = -2
as per the question you have given we cannot find intercepts.
Hope iam clear and right.

Answer 7

There are 2 aspects of a linear equation (the form you're doing) The coefficient and the y intercept. Y=(coefficient)X+(y intercept) So, interior the 1st one, its x+2, 2 is the y intercept and 1x is the coefficient. interior the 2d, a million/2x is the coefficient, and 0 is the y intercept.

Answer 8

To find intercept solve for y=0 for x intercept and x=0 for y intercept

y-1 = x^2-6x+9
y=x^2-6x+10

if x=0
y=0-0+10
y=10

if y=0
0=x^2-6x+10
using quad equation

( -(-6) +/- sqrt [ (-6)^2 - 4(1)(10) ] ) / 2(1)

(6 +/- sqrt [ 36-40] ) / 2
sqrt [ 36-40] <------- equal imaginary number so it does not touch x axis.

Only touches y axis at +10

Answer 9

it easy u plug in a 0 for x then slove and u plug a 0 for y and slove