Question

How do you find the vertex and x intercept of y=4(x-2)^2+1

Answer 1

@maherman97 Do you know Vertex formula?

Answer 2

(h,k) is a vertex point. the equation you have is y=a(x-h)^2+k.
So just substitute the numbers in. For the x-intercept just set the equation equal to zero.

Answer 3

@maherman97 @izaikelly has already explained it so neatly and clearly? Please ask if you need further help ?

Answer 4

I don't understand what to do once you turn the y to zero.

Answer 5

y = a( x - h )^2 + k.
y = 4(x - 2 )^2 + 1

Answer 6

Now, can you see a = ?

Answer 7

yesssirr but i dont know what to do with the equation.

Answer 8

:"find the vertex"

Answer 9

Im sorry, I really dont understand.

Answer 10

(h,k) is a vertex

Answer 11

okay so what do i do to find the x-intercept though?

Answer 12

4(x - 2 )^2 + 1 = 0

Answer 13

4(x - 2) ² + 1 = 0
Can you solve it to find x?

Answer 14

do i do \[4x ^{2}-16x+16\]

Answer 15

Wait, are you sure your post is correct?

Answer 16

yeah the problem is \[ y=4(x-2)^{2}+1\]

Answer 17

you are supposed to graph it too.

Answer 18

when you set y = 0
4(x - 2) ² + 1 = 0
There's no solution for x

Answer 19

4(x - 2) ² = -1
Unless you've learned about complex, imaginary solution!

Answer 20

noo... so what is the x intercept???

Answer 21

The answer is no x intercept means the graph doesn't cut x axis !

Answer 22

OHHH, so i just plug in two more points for the graph?

Answer 23

Yes, if you want to graph it, just plug random values ( choose the one easily to compute)

Answer 24

okay thanks!