Question

y=x^2-4x+8

Find the vertex and x intercepts of the graph.

Answer 1

x-intercept: When y=0
Vertex: 2x-4
2x=4
x=2
The vertex is at x=2, so plug that in and find y

Answer 2

Whats the formula to solve for y then? Would I just switch up the formula or...?

Answer 3

another way is to write it in the form

(x - 2)^2 + 4

Answer 4

@cwrw238 Can finish explaining, I gtg

Answer 5

So, is the formula to solve for y (x-2)^2+4?

Answer 6

this is called the vertex form of the equation
the minimum value of (x - 2^2 is zero ( a square can't be negative) so at the vertex of the graph x = 2 and y = 4

vertex is at ( 2, 4)

Answer 7

the expression has a positive x^2 so will have a minimum value and graph will look like:

Answer 8

there are no intercepts on x-axis

if you try to solve (x - 2)^2 + 4 = 0 what do you get?

(x - 2)^2 = = -4

can u find any real roots?

Answer 9

no?

Answer 10

no - because there are no real roots of a negative number

Answer 11

Okay thank you so much!

Answer 12

yw - have you any more questions about this problem?

Answer 13

Nope