Question

how do i find intercepts of a quadratic equation in this form y=(x-3)squared - 1?

Answer 1

set y to zero and solve for the x intercepts
set x to zero for the y intercepts

Answer 2

the intercepts on the graph are 0,3 and 2,3. can i even get that by setting those to 0?

Answer 3

(2,3) is an intercept?
that's not possible?

unless you meant the intercept with another equation?

intercepts include x intercepts and y intercepts
(2,3) cannot be a x/y intercept

Answer 4

yeah my bad. 2,3 is just the refraction on the otherside of the parabola. but i cant figure out how to get 0,3 from this equation

Answer 5

ok
y=(x-3)^2- 1
set y = 0
(x - 3)^2 - 1= 0
(x - 3)^2 = 1
x - 3 = 1
x - 3 = -1
x = 4
x = 2

i get the x intercept to be (4, 0) and (2, 0)

set x = 0
y = (0 - 3)^2 - 1
y = 9 - 1
y = 8

and the y intercept to be (0, 8)

i put it in a graphing calculator and they're right
so either I don't understand your question, or you typed it wrong

Answer 6

original problem is y-1= (x-3)^2

Answer 7

yeah it's wrong
your equation should have been y = (x-3)^2 + 1

Answer 8

(0, 3) is not a point on the parabola
just try plugging 0 for x it's not possible
you get
y = (0 - 3)^2 + 1
y = 9 + 1
y = 10

(0, 10)

Answer 9

now i see. thankyou. one more question. the vertex on a graph given is 1,1...it gives four equations to choose from but the one it says is the answer is h(x)=(x-1)^2+1. does only the sign before the k value matter?

Answer 10

no
rewrite the equation
y = (x-1)^2 + 1
and again
y - 1 = (x-1)^2

now i'm not sure if i can make myself clear.
the vertex for y = x^2 is (0, 0) yes?
it would be the same if we wrote y - 0 = (x - 0)^2

get it?

the vertex for y - 1 = (x-1)^2 (it shift one unit right and one unit up)
would be (1, 1)