Question

The parabola is a graph of a quadratic equation. What are the roots of the equation?

Answer 1

without knowing the rule of f(x), approximate the x intercepts the best you can

Answer 2

(2,3)

Answer 3

or, if the graph is accurate enough, i see a vertex and a y intercept that can be culled to create a suitable equation from

Answer 4

vertex at -7,3

y = m(x+7)^2 + 3

at x=0, y = 2

2 = m(7)^2 + 3

-1 = 49m

m = -1/49

Answer 5

lol, got that a little out of place

Answer 6

vertex is 3,-7 ....

Answer 7

okay..

Answer 8

vertex at 3,-7

y = m(x-3)^2 -7

at x=0, y = 2

2 = m(-3)^2 - 7

9 = 9m

m = 1

Answer 9

im trying to follow you lol

Answer 10

y = (x-3)^2 - 7 can be solved for y=0

0 = (x-3)^2 - 7

7 = (x-3)^2

+- sqrt(7) = x-3

3 +- sqrt(7) = x

Answer 11

okay.. now what

Answer 12

ok ... show me where this goes awry for you ....

Answer 13

wait what

Answer 14

i assume since i got to the end of it and you are still lost that there must be some place that threw you for a loop; so instead of me doing all the work for you, tell me where this starts to not make sense ....

Answer 15

at this part
0 = (x-3)^2 - 7

7 = (x-3)^2

+- sqrt(7) = x-3

3 +- sqrt(7) = x

Answer 16

thats just solving for y=0; which is how the roots (or zeros) are defined

Answer 17

okay so the answer is (0,3) ??

Answer 18

do you understand how i got the equation?

y = (x-3)^2 - 7 ?

Answer 19

not really lol

Answer 20

sorry im stupid :/ lol

Answer 21

your material should have what is called the "vertex" form of a quadratic equation ... show me what it looks like in its general form

Answer 22

y = a(x – h)2 + k

Answer 23

good, and (h,k) is the point that defines the vertex; what is the point for the vertex on the given graph?

Answer 24

(0,3) ?

Answer 25

not quite .. the vertex is the bend in the graph where it start to go back on itself; it changes directions

Answer 26

-7,3 ?