Question

The square of a number decreased by 25 is -9. How do I find the x intercepts and the vertex of the graph

Answer 1

x intercepts: set equation to zero and solve for x.
vertex: put equation into standard quadratic form and use formula -b/2a, then plug value you get into function and get y coordinate

Answer 2

Can yous how me?

Answer 3

@alanli123

Answer 4

yea

Answer 5

Thanks

Answer 6

\[x ^{2}-25=-9\]

Answer 7

Right then
\[x ^{2} - 25 + 9 = 0\]

Answer 8

right?

Answer 9

What happens next?

Answer 10

Solve for x.

Answer 11

\[x ^{2}=16\]

Answer 12

and then you square them both so x = 4 right?

Answer 13

x = postive AND minus 4

Answer 14

ooh right cuz you squared it, what do we do next?

Answer 15

Wait we kinda skipped something. We have to set the equation to 0, so: \[x ^{2}-16=0\]
Then difference of perfect squares and factor it out. \[(x-4)(x+4)=0\] Solving the equation gives x equals plus or minus 4. (I

Answer 16

We did this because this is the correct approach to solving the equation. The way before was a bit unorthodox.

Answer 17

ohh

Answer 18

So, that means the x intercepts is 4,0 and -4.0

Answer 19

Now, for the vertex. So the quadratic equation we see so far is \[x ^{2}-16 = 0\] Since the coefficient in front of \[x^{2}\] is positive, that means the vertex is a minimum value one. Now, lets think about this with a graphical approach. x^2 - 16 = 0 is just x^2, except 16 units down, right? Well, the minimum value or vertex of x^2 is just 0, so that means if we subtract 16, we obtain the minimum value or vertex of x^2 - 16. So, the vertex of that is 0, -16

Answer 20

Thanks a lot, know your tired (saw it in the chat) so thanks for helping me out

Answer 21

haha np